Question 1203935: At an awards ceremony, nine women and seven men are each to receive an award and are to be presented with their award one at a time. Two of the awards are to be first given to two of the women, and then the remaining awards will alternate between men and women. How many ways can this be done?
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
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The following two statements are contradictory:
"... are to be presented with their award one at a time..."
"Two of the awards are to be first given to two of the women..."
If the awards to the first two women are given one at a time, then the number of ways to do that is 9*8=72; if two of the women are given their awards at the same time, then the number of ways is C(9,2)=36.
After the awards are given to the first two women, the remaining awards are to be given alternating "between men and women". It is not clear from that wording whether the third award must go to a man or if it can go to any of the remaining 14 people. So there are either 7 or 14 ways to choose the recipient of the third award.
After the third award is given, there are 7 people of one gender and 6 of the other. Since the awards must be presented alternating between genders, the number of ways to present awards 4 through 16 is
7*6*6*5*5*4*4*3*3*2*2*1*1 = 3628800.
So, depending on the interpretation of the problem, I see four different possible answers:
(1) 72*14*3628800
(2) 72*7*3628800
(3) 36*14*3628800
(4) 36*7*3628800
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