# [SOLVED] Permutation and Combination Question 3

• Feb 14th 2009, 11:57 PM
azuki
[SOLVED] Permutation and Combination Question 3
11 players are to be selected for a football game In the squad are 3 goal keepers, 6 defenders, 5 mid-fielders and 6 strikers. In how many ways can a coach pick his team if he needs 1 goal keeper, 4 defenders, 3 mid-fielders and 3 strikers?

I get 3C1 * 6C4 * 5C3 *6C3 and get 9000.
Am I right?

In how many ways can 3 red, 1 black, 2 white and 2 blue balls be selected from 6 red, 2 black, 5 white and 4 blue balls; and in how many ways can they be arranged?

I get 6C3 * 2C1 * 5C2 * 4C2
=2,400

For second part of question, I take 8!/(3!2!2!) = 1,680 But the answer sheet shows another answer. Am I wrong, or the answer sheet is wrong?
• Feb 15th 2009, 04:34 AM
Soroban
Hello, azuki!

Quote:

11 players are to be selected for a football game.
In the squad are 3 goal keepers, 6 defenders, 5 mid-fielders and 6 strikers.
In how many ways can a coach pick his team
if he needs 1 goal keeper, 4 defenders, 3 mid-fielders and 3 strikers?

I get: . $\left(_3C_1\right)\left(_6C_4\right)\left(_5C_3\ri ght)\left(_6C_3\right) \:=\:9000$

Am I right? . Yes!

Quote:

In how many ways can 3 red, 1 black, 2 white and 2 blue balls
be selected from 6 red, 2 black, 5 white and 4 blue balls.
And in how many ways can they be arranged?

I get: . $\left(_6C_3\right)\left(_2C_1\right)\left(_5C_2\ri ght)\left(_4C_2\right) \:=\:2,\!400$ . Right!

For second part, I take: . $\frac{8!}{3!\,2!\,2!}\: =\: 1,\!680$

What is it?

What was the original wording of the problem?

"Arranged" is a very sloppy term.

. . $\begin{array}{c}\text{They can be arranged in a row.} \\
\text{They can be placed in a circle.} \\
\text{They can be stacked to form a }2\!\times\!2\!\times\!2\text{ cube.} \\
\vdots \end{array}$