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A television Advertising Break

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by singhsa » Sun Sep 12, 2010 5:21 pm
A television advertising break is to consist of six 30-second advertisements. If the second, fourth, and sixth of the 30-second spots are to be filled with three different advertisements for company X and the other spots are to be filled with one advertisement each for companies A, B, and C, in how many different ways can the six advertisements be ordered?
(A) 729
(B) 720
(C) 120
(D) 36
(E) 24

OA soon
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by GMATGuruNY » Sun Sep 12, 2010 5:36 pm
singhsa wrote:A television advertising break is to consist of six 30-second advertisements. If the second, fourth, and sixth of the 30-second spots are to be filled with three different advertisements for company X and the other spots are to be filled with one advertisement each for companies A, B, and C, in how many different ways can the six advertisements be ordered?
(A) 729
(B) 720
(C) 120
(D) 36
(E) 24

OA soon
Let's call the 6 spots X1, X2, X3, A, B and C.

Number of choices for spot 1 = 3 (A, B or C)
Number of choices for spot 2 = 3 (X1, X2, or X3)
Number of choices for spot 3 = 2 (2 remaining choices of A, B and C)
Number of choices for spot 4 = 2 (2 remaining choices of X1, X2 and X3)
Number of choices for spot 5 = 1 (1 remaining choice of A, B and C)
Number of choices for spot 6 = 1 (1 remaining choice of X1, X2 and X3)

To combine the number of choices for each spot, we multiply:

3*3*2*2*1*1 = 36.

The correct answer is D.
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by Rahul@gurome » Sun Sep 12, 2010 5:36 pm
Let the three different ads from company X be A1, A2, A3.
1st spot can be filled by ad for A or B or C; can be done in 3 ways
2nd spot has to be filled by one of A1, A2, A3; can be done in 3 ways
3rd spot can be filled by 2 remaining ads from A, B, C; can be done in 2 ways
4th spot can be filled by 2 remaining ads from A1, A2, A3; can be done in 2 ways
5th spot has to be filled by 1 remaining ad from A, B, C; can be done in 1 way
6th spot can be filled by 1 remaining ads from A1, A2, A3; can be done in 1 way

Therefore, number of ways in which the six advertisements be ordered = 3*3*2*2*1*1 = 36 ways

The correct answer is [spoiler](D)[/spoiler].
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by singhsa » Mon Sep 13, 2010 9:39 am
GMATGuruNY wrote:
singhsa wrote:A television advertising break is to consist of six 30-second advertisements. If the second, fourth, and sixth of the 30-second spots are to be filled with three different advertisements for company X and the other spots are to be filled with one advertisement each for companies A, B, and C, in how many different ways can the six advertisements be ordered?
(A) 729
(B) 720
(C) 120
(D) 36
(E) 24

OA soon
Let's call the 6 spots X1, X2, X3, A, B and C.

Number of choices for spot 1 = 3 (A, B or C)
Number of choices for spot 2 = 3 (X1, X2, or X3)
Number of choices for spot 3 = 2 (2 remaining choices of A, B and C)
Number of choices for spot 4 = 2 (2 remaining choices of X1, X2 and X3)
Number of choices for spot 5 = 1 (1 remaining choice of A, B and C)
Number of choices for spot 6 = 1 (1 remaining choice of X1, X2 and X3)

To combine the number of choices for each spot, we multiply:

3*3*2*2*1*1 = 36.

The correct answer is D.

Ohh...found out where I went wrong....did not read the portion where it says that the advertisements by company X were different.

So, in case the advertisements by company X were all the same, we would have A,B,C and 3 X's. In this case would the answer be - (3*3*2*2*1*1)/3! = 6 or would there be a different answer to that?
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