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1) A club has 9 women and $8$ men. Count the number of different committees of size 4 with $3$ or $4$ women.

How can I take in account for $3$ women and $1$ man? I have ${9 \choose 4}$ thus far.

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    Well, $\binom 94$ would be the case if all four committee members were women, which is one part of the question. Now, what can you do if there are exactly $3$ women on the committee? – 2017-02-09
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    Count the number of committees with four women. Then count the number of committees with three women. Then add the two. – 2017-02-09

2 Answers 2

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1) For choose $3$ women you have ${9 \choose 3}$ possibilities and for choose $1$ man you have ${8 \choose 1}$, so the total is $${9 \choose 3}\cdot {8 \choose 1}$$

2) For choose $4$ women you have ${9 \choose 4}$ and so the total is $${9 \choose 4}$$

The total is then:

$${9 \choose 3}\cdot {8 \choose 1}+{9 \choose 4}$$

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    My professor said all I was missing is 3 women 1 man?? – 2017-02-09
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    @Arthur: yes, thanks – 2017-02-09
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    can you explain both parts please? – 2017-02-09
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    @RobertMiller: Is it clear? – 2017-02-09
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    Not clear, no. Can you explain each please? Your logic? Why would you account for 0 men? – 2017-02-09
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    @RobertMiller: Actually you don't need that because ${8 \choose 0}=1$. I took it off – 2017-02-09
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    These are all combinations, correct? – 2017-02-09
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    @RobertMiller: yes, they are! because the order doesn't matter – 2017-02-09
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Case 1:

$3$ women and $1$ man

$${}_9 C_3 \times {}_8 C_1 = 672$$

Case 2:

$4$ women and $0$ men

$${}_9 C_4 \times {}_8 C_0= 126$$

Total number of possible committees then is $672+126=798$