The number of combinations is 32!/[6!*(32-6)!] where n! represents 1*2*3*...*n
The answer is 32*31*30*29*28*27/(6*5*4*3*2*1) = 906192
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252 combinations, :)
12,108,327
There are 8,592,039,666 combinations of 6 numbers out of 138 numbers, like the numbers from 1 to 138.
(35!)/(29! 6!) = (35 x 34 x 33 x 32 x 31 x 30)/(6 x 5 x 4 x 3 x 2) = 1,623,160
The sum you require is 4!/2!2!, which is 6. This means there are 6 different two-digit numbers. * * * * * Wrong! The above formula is for the number of combinations, not permutations. However, since the number 23 is different from the number 32, you require permutations and not combinations. There are 4*3 = 12 23, 25, 27, 32, 35, 37, 52, 53, 57, 72, 73, 75.