# Math Help - counting problem

1. ## counting problem

hello!can someone help me solve this:

8 women and 12 men want to form a committee of 5 persons in which there are at least 2 men and 2 women.

1- In how many ways can they form the committee?
2- If Mr. X and Mrs. Y refuse to work together, how many committees can
be formed?
Four married couples have bought 8 seats in the same row for a concert. In how many di erent ways can they be seated:

1- With no restrictions.

2-If each couple is to sit together.

3- If all the men sit together next to the right of all the women.

Thank you

2. ## Combinatorics

Hello qwerty321
Originally Posted by qwerty321
hello!can someone help me solve this:
Originally Posted by qwerty321
8 women and 12 men want to form a committee of 5 persons in which there are at least 2 men and 2 women.

1- In how many ways can they form the committee?
2- If Mr. X and Mrs. Y refuse to work together, how many committees can
be formed?
Four married couples have bought 8 seats in the same row for a concert. In how many di erent ways can they be seated:

1- With no restrictions.

2-If each couple is to sit together.

3- If all the men sit together next to the right of all the women.

Thank you

1 Committee can either have 3W + 2M in $_8C_3 \times _{12}C_2$ ways, or 2W + 3M in $_8C_2 \times _{12}C_3$ ways. Add these answers to get the total number of ways.

2 Work out the number of committees in which the two antagonists will be chosen together - so in how many ways can the remaining 3 be chosen? - and then subtract this from your answer to part 1.

In the second problem:

1 If there's no restriction, this is simply the number of ways of arranging 8 items in a line = 8!

2 If each couple sit together, how many ways are there of arranging the four couples in a line? And in any given arrangement of the four couples, how many ways are there then of seating husband and wife in each couple? Multiply these together to get the overall number.

3 Arrange all four men in a line; then arrange all four women in a line. Multiply the answers.

Can you do it now?

3. can u explain to me more clearly how to solve:
If Mr. X and Mrs. Y refuse to work together, how many committees can
be formed?
thank you

Hello qwerty321
1 Committee can either have 3W + 2M in $_3C_8 \times _2C_{12}$ ways, or 2W + 3M in $_2C_8 \times _3C_{12}$ ways. Add these answers to get the total number of ways.

2 Work out the number of committees in which the two antagonists will be chosen together - so in how many ways can the remaining 3 be chosen? - and then subtract this from your answer to part 1.

In the second problem:

1 If there's no restriction, this is simply the number of ways of arranging 8 items in a line = 8!

2 If each couple sit together, how many ways are there of arranging the four couples in a line? And in any given arrangement of the four couples, how many ways are there then of seating husband and wife in each couple? Multiply these together to get the overall number.

3 Arrange all four men in a line; then arrange all four women in a line. Multiply the answers.

Can you do it now?

2 If each couple sit together, how many ways are there of arranging the four couples in a line? And in any given arrangement of the four couples, how many ways are there then of seating husband and wife in each couple? Multiply these together to get the overall number.

Is it 8!/2!?

3 Arrange all four men in a line; then arrange all four women in a line. Multiply the answers.
4!*4!?

5. Originally Posted by qwerty321
can u explain to me more clearly how to solve:
If Mr. X and Mrs. Y refuse to work together, how many committees can
be formed?
Grandad gave you the total number of possible committees. Call it T.
We find the total number of those that include both X and Y. Call it S
$S={ 11 \choose 1}{ 7 \choose 2}+{ 11 \choose 2}{ 7 \choose 1}$. [Note that we have used X & Y so the numbers are reduced.]
Because we do not want both X and Y on the committee, we remove that number so the answer is $T-S$

6. ok thank you
can you check for me the answers of the other problem?

However, you must do work on #2.
The couples can sit in $4!$ ways.
Each couple can sit in two ways: HW or WH. There are four couples so we have a factor of $2^4$.

8. so it is 4!*2^4..
btw for 1- With no restrictions:

is it 8! or 4! because we are talking about couples right?

9. ## Combinatorics

Hello qwerty321
Originally Posted by qwerty321
so it is 4!*2^4..
btw for 1- With no restrictions:

is it 8! or 4! because we are talking about couples right?
Yes, the number of ways of arranging the four couples is 4!, and each of the 4 couples can sit in 2 ways. So that's $4! \times 2^4$ ways altogether,

If there are no restrictions at all, the answer is 8! - we are not talking about couples, just 8 individuals.