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Bunuel
In how many ways can 11 # signs and 8 * signs be arranged in a row so that no two * signs come together?

A. 108
B. 126
C. 252
D. 495
E. 990
One other approach here could also be to place all 8 * signs and 7 # signs in the middle. The problem now requires us to feel in 9 gap areas around 8 * signs with remaining 4 # signs. Kind of a variant of beggars problem where you have to divide 4 things among 9 variables.

r1 + r2 + r3 + ... + r9 = 4

Solution for such equations is n+r-1Cr-1 = (9+4-1)C(9-1) = 12C8 = 495

IMO: D
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For this question, I kind of grouped the "*#" together and solved it like MISSISSIPPI problem.

I got 11!/8!3! from there and got 165, and then I multiplied it by 2 because the grouping could also be "#*" and got 330. Now of course its not there, but I still answered D because it was the closest value with 165 * 3. However, I'm not sure what exactly I'm doing wrong, or what case I'm forgetting. Can somebody add some input?
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pk14y
For this question, I kind of grouped the "*#" together and solved it like MISSISSIPPI problem.

I got 11!/8!3! from there and got 165, and then I multiplied it by 2 because the grouping could also be "#*" and got 330. Now of course its not there, but I still answered D because it was the closest value with 165 * 3. However, I'm not sure what exactly I'm doing wrong, or what case I'm forgetting. Can somebody add some input?

's and #'s are identical, so arranging those does not make sense. Check the methods described above.
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