# Perms and Combs

1. Jan 17, 2005

### Ryoukomaru

4 children out of 8 will be selected. But two oldest children can not be both chosen. Total number of combinations = ?

The mutually exclusive situations are really confusing me.
I know if they were independent, total n. of combinations would be
$$\frac{8!}{(8-4)!4!}=70$$

I need to subtract the total number of combinations with one of the two oldest boys. I am at a dead end, even though i feel like i know how to do it. My mind kinda went blank.

PS. This is the last question i need to answer to finish my "beutifully done" homework. help :P I have high expectations from this piece of art.

Last edited: Jan 17, 2005
2. Jan 17, 2005

### Gokul43201

Staff Emeritus
You need to subtract the no. of combinations involving both of the oldest boys. How many such combinations will there be ?

Simple. First two pick both of the oldest boys. This can be done in exactly 1 way. Now you have to pick 2 more boys from the remaining 6. You know how to do this.

Subtract from original to get final answer.

3. Jan 17, 2005

### Ryoukomaru

Ahhh that would be $$^6C_2=15$$ so 55 is the final answer. Thx a lot.

I have been looking at the question from a different angle since the beginning.

Last edited: Jan 17, 2005