College algebra?

I'm am in a independent study class, i can't seem to firgure out these questions. They are all related to each other, or are of the same lesson. The first person to answer these questions correctly gets 10 points. I have the answers I just need a better explaination so I understand them better.

1. The mellowmusic CD had nine songs on it. How many ways can you select three of the nine songs to listen to if order doesn't matter? (Find the number of combinations of nine songs taken three at a time.)

2. Dave also plays volleyball. For his team, he needs four men and three women, including substitutes. There are eight men and six women who want to play. In how many different ways can he choose a team?

3. How many seven digit telephone numbers are possible if there are no restrictions on the digits you are allowed to choose?

Great answers, no one has got number three yet.

What I need is a tutorial (step by step instructions).

Thanks so much for helping me out.

1) C(9, 3) = 9x8x7/(3x2x1) = 84 ways.

2) C(8, 4) x C(6, 3)
= 8x7x6x5/(4x3x2x1) x 6x5x4/(3x2x1) = 70 x 20 = 1,400 ways.

3) If you have no restrictions, you have 10 choices for each digit: 10^7 = 10,000,000 ways. This definitely doesn't reflect reality though, cuz you can't have the number 000-0000 for instance.

9x8x7x6x5x4x3x2x1 lol

1. This is the most difficult problem. Every time you select a song you always have one less song to chose. So the total number of combinations of nine songs taken 3 at a time would be 9x8x7=504.

2. This problem is tackled by looking at the total number of combinations of male players and female players.

# Combinations Male Players= 8x7x6x5= 1680

# Combinations Female Players= 6x5x4= 120

For Every Combination of Male Players every combination of Female players is possible. So the total # of combos is 1680x120= 201,600.

3. This is the easiest of the questions. A seven digit phone number with 10 possible number per digits the total number of combination is: 10x10x10x10x10x10x10= 10000000.

There You Go

Mark

1) 9c3= 9! / (3!*6!)
= 362880 / (6 * 720) = 84 ways

2) 8c4 * 6c3 = 8!/( 4! * 4!) * 6! / (3! *3!)
= 70 * 20 = 1400 ways

3) 9999999 numbers

with two digits you can have the numbers 00 until 99 these are 100 numbes.
so with 7 digits you can have 1000000 numbers ( including 0000000 )