Question

A club is comprised of 10 guys and 7 girls. How many different photos can be made that consists of 9 club members in a row with the guys and girls alternating postions?

Answer 1

We have two possible configurations for photos8
1) M-F-M-F-M-F-M-F-M : 5 guys and 4 girls;
2) F-M-F-M-F-M-F-M-F : 5 girls and 4 guys.
How many possibilities for the first case?
You have to think about "combinations" in mathematics and we apply it in these steps:
1) firstly you can choose a guy between the 10 => 10 possible choices;
2) now you can choice a second guy between the 9 they remain => 9 possibilities. In total you have at this step 10x9 = 90 different possibilities!
3) the third guy can be found in the 8 they remain. So you have 10x9x8;
and so on.
So we say: the 5 guys we choose to compose the first configuration represent one of the
10x9x8x7x6 = 30240 possibilities.
In the same way the 4 girls of the first configuration represent one of the 7x6x5x4 = 840 possible choices.
in total we have: 30240 x 840 = 25401600 photo we can make for the first MFMFMFMFM configuration.

Answer 2

Anyway when we make that photo, the order of the components could be not important.
So, for example, how may permutations can I have with a certain number of objects?
For example, 3 objects: abc, acb, bac, cab, bca, cba; 6 permutations.
In general nx

Answer 3

(n) x (n-1) x (n-2) ... x 1 = n! (factorial).
[you can understand this because it is similar to the choice of the guys I told you before].

So if it is not important the order of the components in the photo we have to make a division for:
5! (n. of guys) and 4! (n. of girls) and we obtain (for the first possible type of photo):
25401600 / (120 x 24) = 8820 photos.

Answer 4

Can you tell me how many photos can be made for the second configuration (FMFMFMFMF)?