Finding the Total Combinations When Two Oldest Children Cannot Be Both Chosen

[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.

 

[Gokul43201]

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.

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.

 

[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.