31C5 which is 31!/[5!*(31-5)!] = 31*30*29*28*27/(5*4*3*2*1)
= 169911
31C5 = 169911
we think its 3 comb.
The answer is 31C5 = 31!/[26!5!] = 169,911
There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
There are many different combinations. One example is two and 31.
There are 5245786 possible combinations and I am not stupid enough to try and list what they are!
There are 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
5*4*3*2*1 = 120 combinations. * * * * * No. The previous answerer has confused permutations and combinations. There are only 25 = 32 combinations including the null combinations. There is 1 combination of all 5 numbers There are 5 combinations of 4 numbers out of 5 There are 10 combinations of 3 numbers out of 5 There are 10 combinations of 2 numbers out of 5 There are 5 combinations of 1 numbers out of 5 There is 1 combination of no 5 numbers 32 in all, or 31 if you want to disallow the null combination.
Just one. When considering combinations, the order of the numbers makes no difference. So there is only one combination with those 7 numbers.
A lot