Question 1162947: A committee of 5 members is to be selected from 6 seniors and 4 juniors. Fine the number of ways in which this can be done if the committee has at least 1 junior.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Let's find the number of ways to form a committee without any restrictions. We have 6+4 = 10 people. There are 5 slots to fill.
If order did matter, then we have 10*9*8*7*6 = 30,240 permutations
However, on a committee order doesn't matter. So we divide by 5! = 5*4*3*2*1 = 120 to get 30,240/120 = 252
So there are 252 different committees possible. Let A = 252 as we'll use it later.
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Let's consider only selecting the seniors. We have 6 seniors to pick from and 5 slots to fill. There are 6*5*4*3*2 = 720 different permutations and 720/120 = 6 combinations here. You can also think of this as "there are 6 ways to not pick a senior". Again order doesn't matter.
Let B = 6.
Since there A = 252 ways to form a committee of juniors+seniors, and 6 ways to form a committee of nothing but seniors, this must mean there are A-B = 252-6 = 246 ways to form a committee where there is at least one junior. The phrasing "at least one" means "one or more".
This subtraction works because you basically have two scenarios
scenario 1) the committee is nothing but seniors (ie, no juniors)
scenario 2) the committee has one or more juniors
one or the other must happen and they combine to form the entire set of possibilities or outcomes.
Note: for this problem, you can use the nCr combination formula. To compute the value of A above, use n = 10 and r = 5. To get the value of B, use n = 6 and r = 5.
Another note: the longer method is to compute the number of ways to have 1 junior, 2 juniors, 3 juniors, 4 juniors, and 5 juniors (each separate cases). After all that, you add up those five values. This is a lot more work than necessary.
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Answer: 246
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