Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

share|improve this question
1  
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
2  
3 instead of 11? You must have been reading too many binary numbers! :-) –  celtschk Aug 20 '12 at 16:54

4 Answers 4

up vote 6 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.