In a 10 item mathematics problem solving test how many ways can you select five problems to solve

Consider 10 items (a,b,c,d,e,f,g,h,i,j) There are 10 different options of choosing first number, say a if chosen. This leaves 9 different choices for second, say c is chosen. This leaves 8 different choices for third, say d is chosen. This leaves 7 different choices for fourth, say f is chosen. This leaves 6 different choices for fifth, say h is chosen.

So (a,c,d,f,h) has been selected but total possible was 10x9x8x7x6 which using factorial notation can be written as 10!/5!

Now if the order is not important and only the combination is needed, then (a,c,d,f,h) is the same as (d,a,f,h,c). This means that we need to divide the answer by all the possible patterns (permutations) for five letters. By similar argument the possible number of options for choosing the first is 5, the second is 4 and so on giving the answer to be 5x4x3x2x1 which can be written as 5! So the number of combinations of 5 items from 10 is 10!/(5!5!).

Note 5! Appears twice in our denominator because it was the number used in original question. If we had been choosing 4 from 10 the answer would have been 10!/(4!6!)

General equations are number of permutations (i.e. patterns) of r objects from n objects, all objects being different is n!/r!

General equations are number of combinations of r objects from n objects, all objects being different is n!/((r!(n-r)!)