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4 boys and 5 girls are asked to form a line. In how many ways can this be done?
Find also the number of arrangements in which
a) The first two are girls
b) The first is a boy and the last is a girl
c) The boys are together
d) No two girls stand next to each other
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(_) For the question, not marked by a letter, the answer is (4+5)! = 9! = 1*2*3*4*5*6*7*8*9 = 362,880.
a) (5*4) = 20 different choices for the first two girls,
and (9-2)! = 7! = 1*2*3*4*5*6*7 = 5040 permutations for the rest (9-2) = 7 persons.
Total is (5*4)*7! = 20*5040 = 100,800 different ways.
(b) 4 different choices for the first boy;
5 different choices for the last girl;
and 7! different permutations for the rest of participants.
The total is (4*5)*7! = 20*7! = 100,800 different ways.
(c) 6 choices to place the block of 4 boys in the line;
4! = 24 permutations of 4 boys inside this block
and 5! = 120 permutations for the 5 (five) girls.
In all, it gives 6*24*120 = 17280 different ways.
(d) The scheme is unique: the girls are in positions numbered by odd digits from 1 to 9;
the boys are in positions numbered by even digits from 1 to 9.
Inside this scheme, 5! = 120 permutations are available for girls and 4! = 24 permutations for boys.
In all, 120*24 = 2880 different ways.
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