Guide :

How many numbers can be made with 3 beads and 3 columns, and then 4 beads and four columns

You have 3 beads and 3 columns, how many numbers can you make e.g. 120, 003, 102 etc

Research, Knowledge and Information :


How many different necklaces can be make from 5 red beads and ...


... one with $4$ greens together, four with $3+1$ greens, ... How many necklaces can be made with 6 beads of 3 ... reshaping k columns to 2 columns representing ...
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Beads and numbers: you could make the beads coincide with the ...


Beads and numbers: you could make the beads ... could have the children make their own, then count how many beads ... four, " etc that way kids can work on ...
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combinatorics - How many different necklaces problem ...


How many different necklaces can be made from 20 beads, ... How many different necklaces problem. ... different colored beads (say yellow&green&blue) then I should ...
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6 Beads : nrich.maths.org


If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made ... 6 Beads. If you put three ... three beads and ...
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Adding with the Abacus : nrich.maths.org


Adding with the Abacus. Stage: 1 ... or clay pellets were used to place on the columns to make numbers. ... like the picture above, and then move the beads to add on ...
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Beads and numbers: you could make the beads coincide with the ...


Beads and numbers: you could make the beads coincide with the ... (they'll be marked for how many) then they can put them in order ... and columns on a baking tray to ...
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I love the hands-on concept....Number Bracelets! Students ...


Students write down all of the ways they can make numbers by manipulating the beads on ... show 4 different ways to make the number. Then, ... numbers. Four dice: use ...
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How to Use an Abacus (with Pictures) - wikiHow


Dec 24, 2016 · How to Use an Abacus. ... The abacus should have 3 beads up in the farthest column left, four up in the next ... The number of columns can be whatever you ...
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Explore Teaching Numbers, Math Tubs, and more! - Pinterest


Students write down all of the ways they can make numbers by manipulating the beads ... Teaching numbers / numeros Can ... Can also make a number web and then ...
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Suggested Questions And Answer :


How many numbers can be made with 3 beads and 3 columns, and then 4 beads and four columns

Consider the different combinations of 3 beads: 3 together, 3 all separate, and a bead and a pair of beads. Now the columns. With 3 columns we can choose which column to put the 3 beads together in; that gives us 3 different numbers (003, 030, 300). Separate all the beads, one per column, that's another number (111). That leaves us with a single bead and a pair, i.e., the digits 0, 1 and 2. There are 6 ways of permuting three different objects. Add this to the other 4 and we get 10, so there are 10 numbers possible using three beads and three columns. With 4 beads, all kept together and 4 columns, there are 4 numbers. Separate the beads and put one in each column and we get one more number. That leaves us with two pairs, and a single bead and 3 together. The numbers made of two pairs consist of the digits 0, 0, 2 and 2. If this had been four different digits we would have 24 different permutations, but we only have two different digits, so we need to reduce the permutations by a factor of 2 twice, so 24/4=6. Finally we have the digits 0, 0, 1 and 3. If this had been just 0, 1 and 3, i.e., three different digits it would be similar to the 3 bead problem and there would be 6 ways of permuting these digits. The extra 0 can be at the beginning or end of each of these 6 permutations making 12 altogether. Another way of looking at it is to reduce the 24 permutations of 4 different objects by a factor of 2 to compensate for two digits (0) being the same. So to summarise by adding together all these in order we have 4 (all beads together) + 1 (all beads separate) + 6 (two pairs) + 12 (single bead and 3 beads together) = 23 different numbers.
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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How many rows made up of 5 columns do you get creating combination of 16 numbers, no reaping sets?

For the first number out of 16 there is a choice of 16; for the second there are 15 remaining; then 14, and so on. In total then there are 16*15*14*13*12 but there are 120 ways of arranging these numbers, so as we are only interested in combinations where order does not matter, we need to divide by 120. That gives us 4,368 rows of 5 columns with no repeating sets.
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how many matches

The formula for a square of matches, where there are n small squares along each side, is calculated thus: There are n horizontal matches and n+1 vertical matches to form each row of open squares, where the squares have no base. There are n such rows. The last row is closed (all the squares are given a base) by adding n matches. The formula for the number of matches is N=n(n+n+1)+n=2n^2+2n=2n(n+1). When N=24, 2n(n+1)=24, n(n+1)=12, so n=3. Therefore, you need 24 matches to make a 3X3 square (9 small squares), and 40 to make a 4X4 square, as shown in the table: n n^2 The number of matches, N=2n(n+1) (n>0) 1 1 4 2 4 12 3 9 24 4 16 40 5 25 60 6 36 84 To make a rectangle of aXb squares, that is, a columns and b rows, you need a horizontal matches for each row and a+1 vertical matches. They form a row of a open squares. So you have b(a+a+1) matches for b rows and another a matches to close the last row: N=b(2a+1)+a=2ab+a+b. When a=b, N=2a^2+2a=2a(a+1), as expected. If a=4 and b=10, N=94. Note that if a=10 and b=4, the answer is still the same, N=94, so it doesn't matter which of a and b represents the row and column, because they're interchangeable.
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What is a stem and leaf plot and when would I use one?

Stem and Leaf plots are just a method of ordering data in a dataset to produce a frequency chart. These plots are used in statistical analysis to draw conclusions about a dataset. The usual way this is done is to use part of each datum to create a data bin. Let's imagine a dataset where all the data consists of numbers between 1 and 99. It doesn't matter how big the dataset is or if there are duplicates. Now imagine 10 bins. The first bin is for numbers between 1 and 9; the second for numbers between 10 and 19, and so on. The numbers of the bins will be labelled 0 to 9. The bins are the stems. So we just go through all the data and put each datum into its appropriate bin. But we don't have to put the whole of the data into each bin, because the bin number is already numbered with the first digit of the data. So the contents of each bin just contain the second digit of the data. The bins (stems) are lined up in order 0 to 9 and we can also stack their contents so that the single digits are in order inside the bins. These are the leaves. Imagine the bins are made of glass. We can look at the bins and the heights of the stacks of contents. The heights of the contents form a shape as we run down the line of bins. These heights tell us how many data there are in each bin and indicate where the most data is and where the least data is. This is is a frequency distribution. It's the basis of the Stem and Leaf plot and can be represented by a table or chart. Each row of the table starts with the bin number (STEM) and along the row we have the contents of the bin (LEAVES). Turn the table on its side and we have a chart with the stem running along the bottom and the leaves forming towers over the stems. The chart resembles the row of bins with the stack, or column, of contents over them, but the bins are now invisible, and only their labels remain as regular horizontal divisions on the chart. But it doesn't stop there. This frequency chart tells us where most of the data can be found, where its middle is and the general shape of the data. These are important statistical observations. Not all the bins may have data in them, and some will have lots of data. Random data will produce no particular shape, but in many cases there will be a pattern. We've considered numbers from 1 to 99, but the data can have any range as long as the data is binned carefully to reflect the relative magnitude of the data. If the data were between 250 and 400, for example, we might take the first 2 digits as the bin label: 25 to 40 and the contents would be the third digit. So you need to make a decision based on the range of data values to decide how the data is going to be binned. I hope this helps you to understand Stem and Leaf plots.
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How do I read Stem and Leaf plots?

Stem and Leaf plots are just a method of ordering data in a dataset to produce a frequency chart. The usual way this is done is to use part of each datum to create a data bin. Let's imagine a dataset where all the data consists of numbers between 1 and 99. It doesn't matter how big the dataset is or if there are duplicates. Now imagine 10 bins. The first bin is for numbers between 1 and 9; the second for numbers between 10 and 19, and so on. The numbers of the bins will be labelled 0 to 9. The bins are the stems. So we just go through all the data and put each datum into its appropriate bin. But we don't have to put the whole of the data into each bin, because the bin number is already numbered with the first digit of the data. So the contents of each bin just contain the second digit of the data. The bins (stems) are lined up in order 0 to 9 and we can also stack their contents so that the single digits are in order inside the bins. These are the leaves. Imagine the bins are made of glass. We can look at the bins and the heights of the stacks of contents. The heights of the contents form a shape as we run down the line of bins. These heights tell us how many data there are in each bin and indicate where the most data is and where the least data is. This is is a frequency distribution. It's the basis of the Stem and Leaf plot and can be represented by a table or chart. Each row of the table starts with the bin number (STEM) and along the row we have the contents of the bin (LEAVES). Turn the table on its side and we have a chart with the stem running along the bottom and the leaves forming towers over the stems. The chart resembles the row of bins with the stack, or column, of contents over them, but the bins are now invisible, and only their labels remain as regular horizontal divisions on the chart. But it doesn't stop there. This frequency chart tells us where most of the data can be found, where its middle is and the general shape of the data. These are important statistical observations. Not all the bins may have data in them, and some will have lots of data. Random data will produce no particular shape, but in many cases there will be a pattern. We've considered numbers from 1 to 99, but the data can have any range as long as the data is binned carefully to reflect the relative magnitude of the data. If the data were between 250 and 400, for example, we might take the first 2 digits as the bin label: 25 to 40 and the contents would be the third digit. So you need to make a decision based on the range of data values to decide how the data is going to be binned. I hope this helps you to understand Stem and Leaf plots.
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please answer the question an the information for the question

go part wae at 8mph rest at 4 mph distans=48 & time=7 ours x+y=7 ours, so y=7-x distans=8*x+4*y...8*x+4*(7-x)...8x-4x+28=48....4x=20....x=5 ours y=2 ours
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why in februry there 28 or 29 days and other month 30 or 31?

Please dont put these kinds of questions here anymore but here is the answer i copied it off of an amazon ask.com web sit which you could have just googled for./ ********************************************************************************** To meticulous persons such as ourselves, having the calendar run out in December and not pick up again until March probably seems like a pretty casual approach to timekeeping. However, we must realize that 3,000 years ago, not a helluva lot happened between December and March. The Romans at the time were an agricultural people, and the main purpose of the calendar was to govern the cycle of planting and harvesting. February has always had 28 days, going back to the 8th century BC, when a Roman king by the name of Numa Pompilius established the basic Roman calendar. Before Numa was on the job the calendar covered only ten months, March through December. December, as you may know, roughly translates from Latin as "tenth." July was originally called Quintilis, "fifth," Sextilis was sixth, September was seventh, and so on. Numa, however, was a real go-getter-type guy, and when he got to be in charge of things, he decided it was going to look pretty stupid if the Romans gave the world a calendar that somehow overlooked one-sixth of the year. So he decided that a year would have 355 days--still a bit off the mark, admittedly, but definitely a step in the right direction. Three hundred fifty five days was the approximate length of 12 lunar cycles, with lots of leap days thrown in to keep the calendar lined up with the seasons. Numa also added two new months, January and February, to the end of the year. Since the Romans thought even numbers were unlucky, he made seven of the months 29 days long, and four months 31 days long. ********************************************************************************** But Numa needed one short, even-numbered month to make the number of days work out to 355. February got elected. It was the last month of the year (January didn't become the first month until centuries later), it was in the middle of winter, and presumably, if there had to be an unlucky month, better to make it a short one. ********************************************************************************** Many years later, Julius Caesar reorganized the calendar yet again, giving it 365 days. Some say he made February 29 days long, 30 in leap year, and that Augustus Caesar later pilfered a day; others say Julius just kept it at 28. None of this changes the underlying truth: February is so short mainly because it was the month nobody liked much.
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magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04
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Eight horses are entered in a race...(Have more)

1(a) How many different orders are possible for completing the race? 8*7*6*5*4*3*2*1=8!=40,320 (b) In how many different ways can first, second, and third places be decided? (Assume there is no tie.) (8*7*6)=P(8,3)=336 or 6*C(8,3) where 6=3*2*1 the number of ways of arranging three items 2.)Telephone numbers consist of seven digits; the first digit cannot be 0 or 1. How many telephone numbers are possible? 8*10^6=8,000,000  3.)In how many ways can five people be seated in a row of five seats? 5*4*3*2*1=5!=120 4.)In how many ways can five different mathematics books be placed next to each other on a shelf? 5!=120 5.)In a family of four children, how many different boy-girl birth-order combinations are possible? (The birth orders BBBG and BBGB are different.) 16=2*2*2*2 from BBBB to GGGG 6.)Two cards are chosen in order from a deck. In how many ways can this be done if (a) the first card must be a spade and the second must be a heart?  13*13=169 (b) both cards must be spades? 13*12=156 7.)A company’s employee ID number system consists of one letter followed by three digits. How many different ID numbers are possible with this system? 26*10*10*10=26,000 Continued in comment...
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