Stats question.

[luckwme]

Part 1.
In how many ways can a television director schedule a sponsor's six different commercials during the six time slost allocated to commercials during an hour "special."

I think the answer here is 6*6=36


Part 2
In how many ways can the television director fill the six time slots for commercials if the sponsor has three different commercials, each of which is to be shown twice.

I'm not sure how to approach this for the answer. The answer is 90.

[GmatG]

Originally Posted by luckwme

Part 1.
In how many ways can a television director schedule a sponsor's six different commercials during the six time slost allocated to commercials during an hour "special."

I think the answer here is 6*6=36

This can be done in 6! ways.

Originally Posted by luckwme

Part 2
In how many ways can the television director fill the six time slots for commercials if the sponsor has three different commercials, each of which is to be shown twice.

I'm not sure how to approach this for the answer. The answer is 90.

This can be done in 6!/2!*2!*2!
ie, 90 ways.

[SaGo]

Can you please explain how you get 6!/2!*2!*2!
Especially the denominator?

[luckwme]

I think the formula should be (6!)/(2!*2!*2!)

[Meggieski]

I got the first one but need explanation on second. I kind of see why it is 90 but wouldn't have come up with that on myself. A clear explaination would help very much!

[samal]

Originally Posted by Meggieski

I got the first one but need explanation on second. I kind of see why it is 90 but wouldn't have come up with that on myself. A clear explaination would help very much!

That is because interchanging the position of same type of commercials doesn't generate a new sequence and hence those need to be taken out from the final count.

for example "aA BB CC" and "Aa BB CC" should be counted only once and not twice. Hope this makes it clear.

[GmatG]

Originally Posted by SaGo

Can you please explain how you get 6!/2!*2!*2!
Especially the denominator?

There is a formula for Permutations of n items, some alike

From n items, of which p are alike of one kind, q are alike of another kind, and so on, the number of possible permutations is n!/ (p! * q!)

Comming bak to question:
In how many ways can the television director fill the six time slots for commercials if the sponsor has three different commercials, each of which is to be shown twice.

So, n = 6
and let 3 diffrent commercial be A, B and C (each of which is to be shown twice)

So we can safely say.. that the six required commercials are AA BB and CC


now applying the above stated formula.. p are alike of one kind ie, 2 (for A)
similarly q are alike of another kind ie, 2 (for B)
and r are alike of third kind ie, 2 (for C)

putting everything in formula:
n!/p!*q!*r!...
=> 6!/2!*2!*2! ie, 90.

HTH

[luckwme]

I pulled this question from my college stats book. Figured it'd be a good way to refresh before the test.

The explanation fromthe book Labelled Theorem 1.8

The number of ways in which a set of n distinct objects can be partitioned into k subsets with n1 objects in the first subset, n2 objects in the second subset, .... and nk objects in the kth subset is

(N!)/(n1!*n2!*nk!)


Hope this explanation adds on to what GmatG has stated...

it's amazing how simple the problem is from the solution, yet how complicated it seems when reading the question....

Big thanks for all the help.

Not sure how to source this..here's the book I'm using...

John E. Frend's
Mathematical Statistics 6th Edition
Irwin Miller/Marylees Miller. Husband Wife??? (Just Imagine there discussions over dinner trying to stump one another...)
Copyright 1999 Prentice Hall
ISBN 0-13-123613-X