MHB Find Roles Filled w/ Dale or Margaret: 65 Chars

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The discussion focuses on calculating the number of ways to fill roles in a casting scenario involving Dale, Margaret, and other auditioners. The first question asks for the number of ways to fill roles if exactly one of Dale or Margaret is selected, while the second question seeks the probability of both getting a part. Various permutations were attempted but did not yield the correct results. A comprehensive distribution of possibilities was suggested, showing that the total number of ways to fill the roles is 60,480, with specific counts for scenarios involving neither, both, or just one of the two. It was noted that the children's roles do not significantly impact the calculations, simplifying the problem.
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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.
 
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navi said:
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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