1) (i) Note that the order in which you select the members of the commitee does not matter, so you just need to calculate all the options of choosing 3 men out of 6 and multiply that by all options of choosing 3 women out of 7. This is (6C3)*(7C3) = 6!/3!3! * 7!/3!4! = 6!7!/3!3!3!4!

(ii) First we'll need to pick one men and one women. In how many ways can we do this? We pick one man out of 6, so six ways for that, and for every choice of man we have 7 different women we can pick. So we have 6*7=42 ways of choosing one man and one woman. Now, you do the rest the same way as the earlier case. Can you solve it?

2) (i) we can easily observe that the letters a,b,c will have to be spread between the e's, ie the word will be of the form: e_e_e_e. So we are only interested in how many ways we can arrange a,b,c in three slots. Can you follow from here?

(ii) If you number the seats 1,2,...,20, then this problem is equivalent to simply choosing three numbers out of 20, with order being relevant, ie. for set of three sits that you will choose, there are 3! = 6 ways of arranging the students within these seats. There are 20C3 = 20!/17!3! ways of choosing 3 seats out of 20, and we will multiply that by 3! because the order matters. so 6*(20C3).

3) (i) Four of the ten sticks are defective, so when we start from blank there is a 40% chance that we will pull a bad stick. All sticks being defective means we pulled out 3 defective sticks in 3 successive pulls. In the first pull, the chance of pulling a bad stick is 0.4 as mentioned. In the second, assuming we pulled a bad stick in the first time, there are 9 sticks left and 3 are defective, so 0.33 chance of pulling a bad one, hence there is a 0.4 * 0.33 chance of pulling two bad sticks in a row. Can you complete this up to three defective sticks?

(ii) We want only one defective stick to be pulled out. The probability of taking out a bad stick the first time is 0.4, as mentioned. Now we want no bad stick the next times; the probability of this happening, in a similar fashion to the last excersize, is 0.66*0.625, so the probability of pulling one bad stick, where we pull it in the first time, is 0.4*0.66*0.625. Now calculate the probability of this happening in the two other cases and add them together to get the solution.

(iii) We want none of them to be defective; this is solved in the exact same way as the first one. Can you do this one alone?

4) (i) Getting tails on the second and third toss is irrelevant of the other tosses, since all that matters is that you get tail in this specific toss. Hence P(A) = P(B) = 0.5

For P(C), we want a tails on exactly one coin toss. There are three different ways of this happening: tails 1st, heads 2nd and 3rd; heads 1st, tails 2nd, heads 3rd; heads 1st,2nd and tails 3rd. So we need to calculate the chance of each of these happening and add them.

The probability of tails,heads,heads is 1/2*1/2*1/2 = 1/8. Can you do the rest from here?

(ii) We want three tosses that will be: heads,tails,heads. The chance for each toss is 1/2, so we end up with 1/2^3 = 1/8. I don't know what an independant event is so I can't really answer that :X

It's been three years since I've done any probability (and only high school level at that), so you may want to double-check my answers there.

As for the basic counting problems, I find that simply using common sense is mostly the easiest way of solving the problem. You just need to know what you're counting, and make sure it makes sense.

Hope this helped!