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How many different ways can these roles be filled if exactly one of Dale and Margaret gets a part?

navi

New member
Mar 25, 2018
12
I have been trying every possible process for this question:

For this problem, assume 10 males audition, one of them being Dale, 7 females audition, one of them being Margaret, and 4 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.
1) How many different ways can these roles be filled if exactly one of Dale and Margaret gets a part?
2) What is the probability (if the roles are filled at random) of both Dale and Margaret getting a part?

For number 1, I have tried these permutations and none have worked:

Dale but not Margaret, P(9,2)P(6,1)P(4,2) + Margaret but not Dale, P(9,3)(1)P(4,2)
Dale, P(9,2)P(7,1)P(4,2) + Margaret, P(10,3)P(1)P(4,2) - Both Dale and Margaret,P(9,2)(1)P(4,2)
I also tried both by multiplying P(3,1) to denote the 3 roles Dale could get, but that did not work either.
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I have been trying every possible process for this question:

For this problem, assume 10 males audition, one of them being Dale, 7 females audition, one of them being Margaret, and 4 children audition. The casting director has 3 male roles available, 1 female role available, and 2 child roles available.
1) How many different ways can these roles be filled if exactly one of Dale and Margaret gets a part?
2) What is the probability (if the roles are filled at random) of both Dale and Margaret getting a part?

For number 1, I have tried these permutations and none have worked:

Dale but not Margaret, P(9,2)P(6,1)P(4,2) + Margaret but not Dale, P(9,3)(1)P(4,2)
Dale, P(9,2)P(7,1)P(4,2) + Margaret, P(10,3)P(1)P(4,2) - Both Dale and Margaret,P(9,2)(1)P(4,2)
I also tried both by multiplying P(3,1) to denote the 3 roles Dale could get, but that did not work either.
One thing you might consider, which will give you more practice and greater confidence in your results, is to establish the ENTIRE distribution of possibilities. I'll give you one for free!

All Possibilities: (10 * 9 * 8) * (7) * (4 * 3) = 60,480
Neither: (9 * 8 * 7) * (6) * (4 * 3) = 36,288
Both: 3*(1 * 9 * 8) * (1) * (4 * 3) = 2,592
Margaret Only: (9 * 8 * 7) * (1) * (4 * 3) = 6,048
Dale Only: 3*(1 * 9 * 8) * (6) * (4 * 3) = 15,552
Check: 15,552 + 6,048 + 2,592 + 36,288 = 60,480. We probably missed nothing! :)

Also, having specified the ENTIRE distribution, it may become almost trivial to answer any subsequent question.

P.S. You could simplify your life a little on this problem be realizing the children's roles have nothing to do with the probabilities and almost nothing to do with the counts. By "almost nothing", I mean that each of the results created by ignoring the children's roles is simply multiplied by 12 to accommodate this additional consideration.