I have 10 numbers for example – 1,2,3, 4,5, 6,7,8,9,10 arranged in two sets of five numbers each

I arrange the numbers – in the following format

(1 2 3 4 5 ) and ( 6 7 8 9 10)

A = 1 , A = 6
B= 2 , B= 7
C=3 , C = 8
D= 4 , D= 9
E = 5 , E = 10

TASK – Arrange the numbers into groups of five making sure that EACH combination of five numbers reflects ABCDE

NOTE – The format ABCDE must be strictly adhered to when arranging the numbers in groups of five

Some examples are – (1,2,3,4,5) , (6,7,8,9, 10), ( 1,2,3,9,10) (6, 7, 8, 4,5)

QUESTION- HOW MANY POSSIBLE OTHER ARRANGEMENTS ARE THERE IN THE ABOVE CASE.. AND WHAT ARE THEY?

COULD THIS BE DONE FOR 16 numbers arranged in 2 sets of eight and how would that case be solved ?

Here’s one way to think about it. For (a+b)^10, it means:

Choose a or b, 10 times.

Suppose we want to know the coefficient for a^3 * b^7. The neat thing is that picking 3 elements from 10 to “become a’s” means the other 7 automatically become b’s. They have to, since we want a^3*b^7; every element is used.

So how many ways can we pick 3 items from 10? At first we’d think 10*9*8 — a permutation.

But because we’re multiplying, the order doesn’t matter: Saying element 1, 2, and 3 become a’s (in that order) is the same as saying element 3, 2 and 1 become a’s in that order.

So, we do C(10,3) which is the number of groups of 3 we can pick from 10. Permutations would be important if the order of multiplication mattered, but they don’t in this case so we have to divide by the redundancies.

Hope that helps!