SOLUTION: The judges in ‘Beautiful Baby’ competition have to arrange 10 babies in order of merit. In how many different ways could it be done? Two babies are to be selected to be photogr

Question 1142722: The judges in ‘Beautiful Baby’ competition have to arrange 10 babies in order of merit. In how many different ways could it be done? Two babies are to be selected to be photographed. Show in how many ways can this selection be made?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
10 babies arranced in merit can be arranged in 10! ways.
10! = 3628800 ways.
10! is equal to 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.
do the math and you will get 3628800, as shown above.

2 babies from the group of 10 can be selected in c(10,2) ways.
the general formula for c(n,x) is n! / (x! * (n-x)!).
c(10,2) = 10! / (2! * 8!) = (10 * 9 * 8!) / (2! * 8!) = (10 * 9) / (2 * 1) = 45.

here's some references on factorial.

https://www.mathsisfun.com/numbers/factorial.html

here's a reference on combinations and permutations.
combinations are without order.
permutations are with order.

https://www.mathsisfun.com/combinatorics/combinations-permutations.html

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