Combinations

[hamif]

Hi everyone,
I am just an unmatured mathematician trying to solve this combination problem:

"A class contains 8 men and 6 women and there is one married couple in the class. Find the number of ways a teacher can select a committee of 4 from the class where the husband or wife but not both can be in the committee".

At first, I thought I can solve the problem directly. If the committee must include the husband but not wife then we will have 7 combination 3 and 5 combination 3 possible committee OR if the wife must be in the committee but not the husband then we will have 7 combination 3 and 5 combination 3. But I am not sure of this approach. I need someone to help me out please.

 

[Deleted member 4993]

Hi everyone,
I am just an unmatured mathematician trying to solve this combination problem:

"A class contains 8 men and 6 women and there is one married couple in the class. Find the number of ways a teacher can select a committee of 4 from the class where the husband or wife but not both can be in the committee".

At first, I thought I can solve the problem directly. If the committee must include the husband but not wife then we will have 7 combination 3 and 5 combination 3 possible committee OR if the wife must be in the committee but not the husband then we will have 7 combination 3 and 5 combination 3. But I am not sure of this approach. I need someone to help me out please.

It will be better if we first think of the numbr of committees without any restriction = 14!/[4! * 10!] = 7*13*11 = 1001

Then lets choose 2 people out of 12 (excluding the married couple) = 12!/[2! * 10!] = 6 * 11 = 66

So there are 66 ways to choose 4 people where the "couple" is included.

So there are (1001-66 = ) 935 ways to choose 4 people where the "couple" is not included.

 

[soroban]

Hello, hamif!

A class contains 8 men and 6 women, including one married couple.
Find the number of ways a teacher can select a committee of 4 from the class
where the husband or wife but not both can be in the committee.

I assume that we can have a committee where neither the husband nor the wife is chosen.
. . Then Subhotosh's answer is correct.

We can solve the problem directly . . . if we're careful.


There are 14 in the class: one Husband, one Wife, and twelve Others.


There are three cases to consider:

[1] The Husband is on the committee (and the Wife is not).
. . Then we must choose 3 of the 12 Others.
. . \(\displaystyle \text{There are: }\,_{12}C_3 \:=\:\frac{12!}{3!\,9!} \:=\:220\text{ ways.}\)

[2] The Wife is on the committee (and the Husband is not).
. . Then we must choose 3 of the 12 Others.
. . \(\displaystyle \text{There are: }\,_{12}C_3 \:=\:220\text{ ways.}\)

[3] Neither the Husband nor the Wife is on the committee.
. . Then we must choose 4 of the 12 Others.
. . \(\displaystyle \text{There are: }\,_{12}C_4 \:=\:495\text{ ways.}\)


\(\displaystyle \text{Therefore, there are: }\:220 + 220 + 495 \:=\:935\text{ ways.}\)