# 2 probability questions?

How many different outcomes are there in a "2 out of 3" playoff, a 3/5, and a 4/7. I get that the last game has to be won by the winner and the rest of the games the order does not matter. But i do not know how to set it up.

The other question is, Find the coefficient of x^3y^2z^3w in expansion of (2x+3y-4z+w)^9

i have 9! / ( 8! 6! -64! 1!) but im not sure if thats right.

Please explain and give any help you can. Thanks!

### 1 Answer

- michaelLv 79 years agoBest Answer
You can draw a tree diagram for the first and stop each path once you have a winner. For a 2 of 3 series with teams A and B the possibilities are AA, ABA, ABB,BAA,BAB,BB for 6 outcomes. (all but AAA, BBB).

For a 3 of 5 series you would exclude any of the 32 paths with 4 or 5 As or Bs.

32-5-5-1-1=20 outcomes are possible. The outcomes with A winning are: AAA,AABA,AABBA,ABAA,ABABA,ABBAA,BAAA,BABAA,BBAAA,BAABA.

For a 4 of 7 series A can win 4 straight (1 outcome), or 3 of 4 and the last (4 outcomes) or 3 of 5 and the last (10 outcomes) or 3 of 6 and the last (20 outcomes). So A can win 35 ways and B can win 30 ways. 70 outcomes total.

For the second problem I get [9! / (1!*3!*2!*3!) ] * 2^3 *3^2 *(-4)^3. The fraction [9! / (1!*3!*2!*3!) ] is the number of terms with the specified exponents. The product 2^3 *3^2 *(-4)^3 is due to the coefficients of the original polynomial raised to the power on the associated variable. The result is -23,224,320.