# Ways of selecting at least one man

We have to select a group of 7 out of a group of 9 men and 11 women

Q : How many seven member teams consist of at least one man ?

Now I know that the answer is ${20 \choose 7}-{11 \choose 7} = 77190$

But my first answer was ${9 \choose 1}{19\choose 6}=244188$; because we have to select one man and the rest can be either men or women. I know this is wrong but I don't know why. Any help will be appreciated.

EDIT : Sorry the question should be 9 men and 11 women ( not 3)

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I suppose that "all of them" is not an acceptable answer? – MJD Aug 20 '12 at 16:46
Is there any possibility that the number of women is $11$ !! – Saurabh Aug 20 '12 at 16:49
@SaurabhHota Sorry,see my edit above – Daya Wickremasinghe Aug 20 '12 at 16:51
3 instead of 11? You must have been reading too many binary numbers! :-) – celtschk Aug 20 '12 at 16:54

Your answer of $\binom91\binom{19}6$ counts many of the allowable groups more than once. Suppose that the men are $M_1,\dots,M_9$, and the women are $W_1,\dots,W_{11}$. Consider, for instance, the group consisting of $M_1,M_2,M_3,M_4,W_1,W_2$, and $W_3$: it gets counted four times, once for each of the four men in it. You count it once when you count $M_1$ as the one man counted by the $\binom91$ factor; you count it once more when you count $M_2$ as that man; and you count it yet one more time for each of $M_3$ and $M_4$. In fact, every $7$-person group gets counted once for each man in it, and this results in massive overcounting.

The easiest way to get the right answer is to to start with the $\binom{20}7$ possible $7$-person groups and subtract the $\binom{11}7$ all-woman groups to get a final result of $\binom{20}7-\binom{11}7$ groups containing at least one man.

Added: A much more long-winded way is to count separately the groups with one man, the groups with two men, and so on. When you choose a group with $m$ men and $7-m$ women, you can choose the $m$ men in $\binom9m$ different ways and the $7-m$ women in $\binom{11}{7-m}$ different ways. Thus, by the product rule you can form such a group in $\binom9m\binom{11}{7-m}$ ways. The total number of groups containing at least one man is therefore $$\sum_{m=1}^7\binom9m\binom{11}{7-m}\;,$$ and it’s certainly possible to compute all $7$ terms and sum them. But it’s ever so much easier just to take the total number of possible $7$-person groups and subtract the easily calculated number that include no men.

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Now this is what I call an answer. Appreciate your help. – Daya Wickremasinghe Aug 22 '12 at 8:55

You are trying to count the number of committees with at least one man. What you are counting with the $\binom{9}{1}\binom{19}{6}$ is something else.

Imagine (if one can still imagine such a thing) that you are trying to count the number of ways of forming a committee of $7$ if the rules say that the Chair must be male, with the sexes of the rest of the people unspecified. Then $\binom{9}{1}\binom{19}{6}$ would be the right answer.

But with an undifferentiated committee, we are overcounting by quite a bit. What you are counting with your $\binom{9}{1}\binom{19}{6}$ is the set of all ordered pairs $(x,y)$, where $x$ is any man, and $y$ is any bag of $6$ people that does not contain $x$.

So for example, $x=$ Charlie, $y=\{$Dave, Bob, Mary, Mary-Ann, Jane, Janet$\}$ is one of the ordered pairs $(x,y)$ counted in your $\binom{9}{1}\binom{19}{6}$, but so is $x=$ Dave, $y=\{$Charlie, Bob, Mary, Mary-Ann, Jane, Janet$\}$. However, these result in the same committee of $7$.

And it is messy to compensate for the overcounting, because the amount of overcounting depends on the number of men chosen.

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very clear. Thanks – Daya Wickremasinghe Aug 20 '12 at 17:20

If there are a group of 9 men and 3 women(original question), the answer should be as follows:

As there are only 3 women, we must take at least 4 men.

So, the combinations can be (3w,4m),(2w,5m),(w,6m),(0,7m)

So, answer should be $^3C_3\cdot^9C_4+^3C_2\cdot^9C_5+^3C_1\cdot^9C_6+^3C_0\cdot^9C_7$

If there are a group of 9 men and 11 women, the answer should be as follows:

Out of 20 men and women 7 member group can be chosen in $^{20}C_7$ ways.

If we don't allow men, then we choose 7 member group in $^{11}C_7$ ways.

The difference of the last two combinations is the possible number combination of 7 member group which has at least one man as its member.

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Understand what is wrong in your formula with this simple example. men are A1 , A2 , A3 and women are B1,B2,B3 . We have to select 3 person with at lest one men . Here we are selection is based on your formula ,selecting one man from A1,A2 ,A3 and rest are from {B1 , B2 , B3 }+ {A1,A2,A3}-{selected man}.

now see this selection,

A1 from man section, A2 and B1 from the rest selection I= {A1,A2,B1}

take another selection, A2 from man section, A1 and B1 from the rest

selection II ={A2,A1,B1}

Sadly , both selection I and II are same.It is not acceptable formula.

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