Guide :

How many five digit whole numbers can you make useing the numbers 0,1,and 2

The only rule is that you can only use 0,1, and 2

Research, Knowledge and Information :


How many 5 digit numbers can be formed using the digits 1, 2 ...


This number can be either 1, 2 or 3. The obtained 5 digit number will have 3 digits of one kind ... How many 4 digit numbers can be formed by using 0, 1, 2, 3, 4, 5 ...
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Permutations and Combinations Problems - analyzemath.com


Permutations and Combinations Problems. ... How many 2 digit numbers can you make using ... (We now understand the need to define 0! = 1) Example 5: How many 3 letter ...
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How many 5-digit whole numbers can be made with the number 0 1 2?


... and Arithmetic How many 5-digit whole numbers can be made ... whole numbers can be made with the number 0 1 2? ... a 4-digit number (although you could use ...
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How many 3-digit numbers can be formed using only 2 ... - Quora


How many 3 digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if repetitions of digits are allowed?
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How many four digit numbers can be made using 1-2-3-4


... 3-digit_numbers_can_be_made_using_the_digits_2_3_7 ... number can start with 0, and you cannot use ... many four-digit numbers can be formed using 0 1 ...
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How many combinations can you make with six digits ...


How many combinations can you make with ... the greatest whole number with five digits or the ... What is the greatest nine-digit number you can make using nine ...
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How many even 6 digit numbers can be written using the digits ...


How many even 6 digit numbers can be written using ... The rest of the number can use any of the 5 ... how many $7$ digit numbers can be formed using $1,2,3,4,5 ...
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How many 4-digit whole numbers are there? - jiskha.com


can you produce a 2-digit whole number by ... (1,2,3,4,5) how many 3-digit numbers can be formed ... from the digits 0,1,2,3,4,5,6,7,8,9 find the number ...
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Suggested Questions And Answer :


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 many five digit whole numbers can you make useing the numbers 0,1,and 2

INTEGERS anser=3^5 =243
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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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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how many 4 digit numbers have no repeat digits, do not contain 0, and have a sum of digits equal to 28?

Let's see how many ways 4 digits can add up to 28. We start with the highest digit 9 and subtract it from 28: 28-9=19. Now we subtract the next highest digit 8: 19-8=11. The 2 remaining digits must add up to 11: 5+6, 4+7. So we have two numbers to start with: 5689 and 4789. But we can rearrange each of these in 24 different ways. That gives us 48. In fact, the largest number we can make with the digits is 9876 whose digits sum to 30. So to sum to 28 we have only the digits 9, 8, 6 and 5 and 9, 8, 7 and 4. Nevertheless, we'll look at a general method for solving the problem. So that means we only need to work out 4-digit numbers where the digits are in order and add up to 28. When we've found out how many there are we simply multiply by 24 to find out how many 4-digit numbers there are in total.  The tree diagram is supposed to help you work out the 4-digit numbers whose digits sum to 28. It shows how we can take the sum of two pairs of digits and add their sums. For example, 28=11+17, shown at the top of the diagram. Then it shows how 11 and 17 can be the sum of single digits. On the extreme right (the leaves of the tree) you can see what digits make up the sums. We need the leaves that have no repeated digits. At the top of the diagram we have 3 alternative leaves for the sum of 11: 3+8, 4+7 and 5+6. And for 17 we have only 8+9. So  we can't have 3+8 with 8+9 because 8 is repeated, but we can have both of the alternatives. That gives us 4789  and 5689. Using this method we can find any others that work. But we always come up with the same set of digits, and we identified them earlier. So there are 48 numbers (rearrangements of these numbers).
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using 0-9, how many 4 digit sets can be formed, all digits can be repeated, ie 2576, 9471,

If you mean digits can be repeated in the four digit number and all digits from 0 to 9 are available, then numbers can range from 0000 to 9999, that's 10,000 different numbers. If different digits must make up the numbers and permutations are significant, in other words, the order of the digits is important, then after picking the first digit we have 9 digits left to choose from; after picking the second digit we have 8 left to choose from, and so on. Therefore, the calculation is 10*9*8*7=5040 for 4-digit numbers.
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multiply 3.902x23.1

Ignore the decimal points for the time being. Use long multiplication to multiply 3902 by 231:          3902            231          3902      11706      7804            901362 Now, count how many digits there are after the decimal points in both numbers: 3 on the first and 1 on the second, making 4. Count 4 digits from the end of the product and insert the decimal point: 90.1362. Another easy way to work out where to put the decimal point is to estimate the answer by ignoring the decimal point and just taking the whole number parts of the numbers: 3*23=69 or 4*23=92. The answer is going to be somewhere in between these two numbers, so we know the point goes after 90.
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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how many combinations can be made with three numbers

If by numbers the question means digits then there are 6 ways of combining 4, 6 and 8: 468, 486, 648, 684, 846, 864; but the digits can be combined in pairs: 46, 64, 48, 84, 68, 86. They can also be combined by inserting operators (+, -, *, ÷) between them. Taking + and - only we have: 4+6+8, 4+6-8, 4-6+8, 4-6-8, 6+4+8, 6+4-8, 6-4+8, 6-4-8, etc. This makes 4 combinations of numbers and signs for each of the 6 combinations of 3 digits, making 24. We can similarly add another 24 using multiply and divide; then 24 combinations for add with multiply; 24 for add with divide; 24 for subtract with multiply; 24 for subtract with divide. In all we have 16*6=96 combinations using 4 operators. We could go on with the pairs using all these operators  between the pairs. We also have the exponent operator. That makes 5 operators, introducing a factor of 32 for the combinations in place of 16. For pairs of numbers we can only have one operator between them, so the 6 pair combinations produce 6*5=30 arithmetic results. The sums, differences, products, quotients and exponentials will not necessarily be unique (for example 4+6+8=6+8+4), but that's because of the properties of numbers and rules of arithmetic, rather than arranging the numbers in any sort of order.
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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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