Probability combination question (quick 10 points will vote asaic)

A club consists of 11 men and 13 women in how many ways can it select a 5 person committee containing at least two people of each sex

Correct me if im wrong, i just wanna make sure im doing it right

So total no. of ways 24C5

No of ways with no woman = 11C5
No of ways with no man =13C5
No of ways with exactly 1 man= 11C1*3C4
No of ways with exactly 1 woman = 13C1*11C4

No ways of at least 2 men and two women =

total no of ways - (No of ways with no woman + No of ways with no man
+ No of ways with exactly 1 man +
No of ways with exactly 1 woman )

Thanks in advance!!!

I know the number of ways for at least 1 man and at least 1 woman

is 24C5 -(13C5+11C5) [total minus the no of ways with no man minus the number of ways with no woman]

but im not sure if i im doing it right with (at least 2 for each sex)

As Josh mentioned, your first ones are all correct except for the typo.

As for "at least 2 of each sex", yes - this is exactly right! But there's also another way you can calculate it:

11C2 * 13C2 * 20C1 / 3

(Because the extra person will make each combination count 3 times this way.) These answers SHOULD be identical.

24C5 - 11C5 - 13C5 - 11C1*13C4 - 13C1*11C4 = 28600
11C2 * 13C2 * 20C1 = 85800/3 = 28600

You have the right idea. Be careful. No ways with exactly 1 man is incorrect above, prob typo.

11C1*13C4 not 11C1*3C4

I find these much easier to conceptualize if I use the probabilities and multiply those by the total number of options.

That's just the way my brain works, though.