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If m>n, in how many different ways can m men and n women be seated in a row so that no two women sit together?
(A) m!/n!(m-n)!
(B) m!(m+1)!/(m-n+1)!
(C) m!n!/n!(m-n)!
(D) m!(m+1)!/n!(m-n)!
(E) m!n!/(m-n+1)!
Solution:
We can use suitable numbers for m and n to figure out this problem. We can let m = 4 and n = 2 and let the four men be A, B, C, D and the two women be 1 and 2. If there are no restrictions, there are 6! = 720 arrangements since there are 6 people. Let’s now figure out the number of seating arrangements for the restriction (i.e., let’s suppose the two women must sit together). We can have:
ABCD12, ABC12D, AB12CD, A12BCD, and 12ABCD
However, for each the seating arrangements listed above, there are 4! x 2! = 24 x 2 = 48 arrangements since we can arrange the 4 men 4! ways and the 2 women 2! ways. Therefore, there are 5 x 48 = 240 arrangements if the two women must sit together. This means that there are 720 - 240 = 480 arrangements where no two women sit together. Now, let’s see which of the given choices give us the value of 480 when m = 4 and n = 2.
A) m!/[n!(m-n)!] = 4!/[2! x 2!] = 24/4 = 6 → This is not 480.
B) m!(m+1)!/(m-n+1)! = 4! x 5!/3! = 24 x 120/6 = 480 → This is 480. If none of the other choices yields the value of 72, then B must be the correct answer.
C) m!n!/[n!(m-n)!] = 4! x 2!/[2! x 2!] = 24 x 2/4 = 12 → This is not 480.
D) m!(m+1)!/[n!(m-n)!] = 4! x 5!/[2! x 2!] = 24 x 120/4 = 720 → This is not 480.
E) m!n!/(m-n+1)! = 4! x 2!/3! = 24 x 4/6 = 16 → This is not 480.
Answer: B