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

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