Question 1114788: Hello, thank you for helping me with this math question
Consider a class with 12 girls and 10 boys (22 in total):
In how many ways can a committee of 5 consisting of 3 girls and 2 boys be chosen?
a) 0.03
b) 792
c) 0.3759
d) 9900
Here is my thought process which led to none of the given answer choices.
We can easily eliminate A and C, as they are not numbers of different arrangements, rather, they are probabilities, which is not what the question is asking.
I recognized the nature of the question to be a permutation, as you care about which order you are placing the 12 girls in 3 places on the committee. For example, Girl A, B, C is a possible permutation, but B, A, C is also a different permutation that must be counted. The same is true for putting the 10 boys into the 2 other spaces on the committee. So, I tried (12 nPr 3) + (10 nPr 2), which made logical sense in my head as you want 12 girls in 3 different ways on the committee plus the 10 boys you want in 3 different ways on the committee.
This way gets you to an answer of 1410, which is way off from the answer choices,so I was wondering what I did wrong in this thought process.
Thank you for your time and free advice :)
Found 3 solutions by MathLover1, greenestamps, ikleyn:
Answer by MathLover1(20850)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
(1) It is NOT a permutation; on a committee the order is NOT important.
(2) For each of the ways of choosing 3 of the 12 girls there are many ways of choosing 2 of the 10 boys. That "for each" means you need to multiply the two numbers together -- not add them.
Fix those two parts of your analysis of the problem and you should reach answer d.
Answer by ikleyn(52797)  (Show Source):
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