The number of 4 different book combinations you can choose from 6 books is;6C4 =6!/[4!(6-4)!] =15 combinations of 4 different books.
To calculate the number of ways the letters in the word "pencil" can be rearranged, we first determine the total number of letters, which is 6. Since there are two repeated letters (the letter 'e'), we divide the total number of letters by the factorial of the number of times each repeated letter appears. This gives us 6! / 2! = 360 ways to rearrange the letters in the word "pencil."
720 is the number of ways to combine three known letters and three known numbers.For example, the letters A, B & C and the numbers 1, 2 & 3. The total combinations of these 6 characters is:(6 options)*(5 options)*(4)*(3)*(2)*(1) = 720.However, if the three numbers and three letters are unknown and any number or letter is possible, and repeated numbers or letters are acceptable (such as with a license plate), then the total possibilities for each "space" are multiplied together:(26 possibilities)*(26 possibilities)*(26 possibilities)*(10 possibilities)*(10 " ")*(10 " ") = 17,576,000 combinations.That is, there are 26 letters in the alphabet and 10 numbers (0 thru 9).This is assuming that three of the six spaces spaces are reserved for letters and three spaces are reservedfor numbers.If the combination can be any three letters and any three numbers where different combinations are made by changing whether each space contains a number or a letter, then the answer becomes a product and sum of different choose functions and is much more complicated...
360. There are 6 letters, so there are 6! (=720) different permutations of 6 letters. However, since the two 'o's are indistinguishable, it is necessary to divide the total number of permutations by the number of permutations of the letter 'o's - 2! = 2 Thus 6! ÷ 2! = 360
Try 151,600! Permutations & Combinations. P(n,r)=n!(nār)! not P(n,r)=n!/(n!-r!)r! ?
there are 36 different combination possibilities. Try them all.
The first digit can be one of five integers (1 to 5) The second - fourth digits one of six integers (0 to 5). So: number of (valid) combinations is 5*6*6*6=1080 * * * * * This gives the number of permutations, not combinations. I am working on the latter.
20
Say you have the letters A,B, and C. Here are all the possible combinations. * ABC * ACB * BAC * BCA * CAB * CBA So, 6 if you don't repeat any of the letters. If you DO repeat letters, then simply take the number of letters you have, (3 for instance), and multiply it to the power of the number of letters you have. So, for 3 letters, the formula would be 33 . Or if you had 4 letters it would be 44 and so on.
6
256 iThink * * * * * It depends on combinations of how many. There is 1 combination of 4 letters out of 4, 4 combinations of 3 letters out of 4, 6 combinations of 2 letters out of 4, 4 combinations of 1 letter out of 4. Than makes 15 (= 24-1) in all. Well below the 256 suggested by the previous answer.
252 combinations, :)
you can make 6
There are 6C3 = 20 such combinations.
just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436
Number of combinations = 45C6 = 45!/6!(45-6)! = 8,145,060
Could you generate a complete set of 6 number combinations from 45 numbers ?