# combinations

• Jul 28th 2009, 03:49 AM
flyinhigh123
combinations
A commitee of 6 is to be selected from 10 people of whom A and B are two. HOw many commitees can be formed excluding A if B is included

i have already solved the first part of this qs which was if A and B were both included and that came out to be 70, using combination formula
i really need help for the second part. the answer is 140 but i dont know how to get there ><""

and can someone help me with second half of question too?
In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women? What proportion of these committees contain a particular man and 2 particular women?
ive worked out the first part which is 700 but i dont know the second half which is 1/7 T.T

thanks guyz for any help given !
• Jul 28th 2009, 04:53 AM
mr fantastic
Quote:

Originally Posted by flyinhigh123
A commitee of 6 is to be selected from 10 people of whom A and B are two. HOw many commitees can be formed excluding A if B is included

i have already solved the first part of this qs which was if A and B were both included and that came out to be 70, using combination formula
i really need help for the second part. the answer is 140 but i dont know how to get there ><""

[snip]

A committee either contains B or it doesn't. If it does, then you're not allowed to have A. If it doesn't, then you can any of the remaining nine:

(Number of committees with B and not A) + (number of committees without B) = $\displaystyle {8 \choose 5} + {9 \choose 6}$.

Quote:

Originally Posted by flyinhigh123
[snip]
In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women? What proportion of these committees contain a particular man and 2 particular women?
ive worked out the first part which is 700 but i dont know the second half which is 1/7 T.T

thanks guyz for any help given !

Similar to the other question. You need another 2 men from the remaining 5 and another 2 women from the remaining 5. So the total number is $\displaystyle {5 \choose 2} \cdot {5 \choose 2}$. Now divide this number by 700 to get the required proportion.
• Jul 28th 2009, 06:14 AM
Soroban
Hello, flyinhigh123!

Quote:

A commitee of 6 is to be selected from 10 people of whom A and B are two.
How many commitees can be formed excluding A if B is included?

There are ten people: .$\displaystyle \{A,B,C,D,E,F,G,H,I,J\}$

With no restrictions, there are: .$\displaystyle {10\choose6} \:=\:{\color{blue}210}$ possible committees.

But $\displaystyle A$ and $\displaystyle B$ will not serve together.
We will count the committees in which $\displaystyle A$ and $\displaystyle B$ are together.

Since $\displaystyle A$ and $\displaystyle B$ are already on the committee,
. . we must choose 4 more people from the remaining 8 people.
There are: .$\displaystyle {8\choose4} \:=\:{\color{blue}70}$ committees with both $\displaystyle A$ and $\displaystyle B.$

Therefore, there are: .$\displaystyle 210 - 70 \:=\:140$ committees
. . which do not have both $\displaystyle A$ and $\displaystyle B.$

This agrees with Mr. F's solution.

Quote:

In how many ways can a commitee of 3 men and 4 women be chosen from 6 men and 7 women?
What proportion of these committees contain a particular man and 2 particular women?

This is Mr. F's solution . . . with baby-talk.

You are right: there are 700 possible committees with 3 men and 4 women.

There are six men: .$\displaystyle \{A,B,C,D,E,F\}$
and seven women: .$\displaystyle \{T,U,V,W,X,Y,Z\}$

Suppose man $\displaystyle A$ and women $\displaystyle T, U$ must be on the committee.

We must select 2 more men from $\displaystyle \{B,C,D,E,F\}$
. . . and 2 more women from $\displaystyle \{V,W,X,Y,Z\}$

There are: .$\displaystyle {5\choose2}{5\choose2} \:=\:10\cdot10 \:=\:100$ ways.

So the proportion is: .$\displaystyle \frac{100}{700} \:=\:\frac{1}{7}$

• Jul 31st 2009, 08:32 PM
flyinhigh123
thank you so much for your help !
special thanks to Soroban for your clear and step by step explanation, you really helped me =D