Thread: Possibilities? (I don't remember the exact word)

1. Possibilities? (I don't remember the exact word)

Hi! First time poster.

So I'm doing some homework, due in six hours. I'm almost done, but there's a couple of questions I'm not too sure of: one has to do with possibilities, the other not so much.

1. The easier one:
What is the coefficient of X^11 in (3 + X)^15?

I'm guessing it's 15!/11!4! = 1365 but that 3 in (X + 3)is making me uncomfortable.

The one that's bothering me:
2. Suppose that a class contains 12 men and 17 women.
(a) How many different 10 person teams can be made?
(b) How many different 5 person teams can be made that contain at least two men?

(a) Is it 19^10? That's the only thing I can think of (19 = 12 + 17) but for some reason I don't feel right about it.
(b) "At least". That "at least" changes everything. Any help would be appreciated.

2. Originally Posted by Dorcon
Hi! First time poster.

So I'm doing some homework, due in six hours. I'm almost done, but there's a couple of questions I'm not too sure of: one has to do with possibilities, the other not so much.

1. The easier one:
What is the coefficient of X^11 in (3 + X)^15?

I'm guessing it's 15!/11!4! = 1365 but that 3 in (X + 3)is making me uncomfortable.

The one that's bothering me:
2. Suppose that a class contains 12 men and 17 women.
(a) How many different 10 person teams can be made?
(b) How many different 5 person teams can be made that contain at least two men?

(a) Is it 19^10? That's the only thing I can think of (19 = 12 + 17 Mr F butts in: You're kidding, right?) but for some reason I don't feel right about it.
(b) "At least". That "at least" changes everything. Any help would be appreciated.

1. The general term is $\displaystyle {15 \choose r} 3^{15 - r} x^r$. You require the coefficient of $\displaystyle x^{11} \, ....$

2. (a) $\displaystyle {29 \choose 10}$.

(b) $\displaystyle {12 \choose 2} \cdot {17 \choose 3} + {12 \choose 3} \cdot {17 \choose 2} + {12 \choose 4} \cdot {17 \choose 1} + {12 \choose 5} \cdot {17 \choose 0}$.

3. OK, excellent. Thanks.

I don't suppose the answer to the first question would be 110565 would it?

4. Binomial Coefficients

Hello Dorcon
Originally Posted by Dorcon
OK, excellent. Thanks.

I don't suppose the answer to the first question would be 110565 would it?
It certainly would!