# Math Help - Combination/Permutation Problem

1. ## Combination/Permutation Problem

A television director is scheduling a certain sponsor's commercials for an upcoming broadcast. There are six slots available for commercials. In how many ways may the director schedule the commercials?

a) If the sponsor has 6 different commercials, each to be shown 1 time?
(This one I know is 6!)
b) If the sponsor has 3 different commercials, each to be shown 2 times?
From here onwards I don't know how to approach the question.
c) If the sponsor has 2 different commercials, each to be shown 3 times?

d) If the sponsor has 3 different commercials, the first of which is to be shown 3 times, the second 2 times and the third 1 time?

Can anyone please explain how to answer this question, thanks.

2. Hello, ibetan!

A television director is scheduling a certain sponsor's commercials
for an upcoming broadcast. There are six slots available for commercials.
In how many ways may the director schedule the commercials?

a) If the sponsor has 6 different commercials, each to be shown 1 time?

(This one I know is 6!) . Good!

b) If the sponsor has 3 different commercials, each to be shown 2 times?
Call the commercials: . $\{A,A,B,B,C,C\}$

Then there are: . ${6\choose2,2,2} \:=\:90$ ways.

c) If the sponsor has 2 different commercials, each to be shown 3 times?
Call the commercials: . $\{A,A,A,B,B,B\}$

Then there are: . ${6\choose3,3} \:=\:20$ ways.

d) If the sponsor has 3 different commercials, the first is to be shown 3 times,
the second 2 times and the third 1 time?
Call the commericals: . $\{A,A,A,B,B,C\}$

Then there are: . ${6\choose3,2,1} \:=\:60$ ways.

3. Originally Posted by Soroban
Hello, ibetan!

Call the commercials: . $\{A,A,B,B,C,C\}$

Then there are: . ${6\choose2,2,2} \:=\:90$ ways.

Call the commercials: . $\{A,A,A,B,B,B\}$

Then there are: . ${6\choose3,3} \:=\:20$ ways.

Call the commericals: . $\{A,A,A,B,B,C\}$

Then there are: . ${6\choose3,2,1} \:=\:60$ ways.

I dont get why to use combinations? Can you explain please.
the answer to b) 6C4*4C2=90

I found the same answers using permutations n!/a!b!c!

Edit: I used left side equals right side and they are equal equations. But I dont understand the reasoning behind using combinations for this question.

4. [QUOTE=ibetan;470061]I dont get why to use combinations? Can you explain please. the answer to b) 6C4*4C2=90[QUOTE]
Those are not combinations. The notation $\binom{N}{a,b,c}$ is rarely used in textbooks today.
In stands for permutations with repetitions $\binom{N}{a,b,c}=\frac{N}{a!b!c!}$.