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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?

Mathematics
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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.
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
(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.

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

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