SOLUTION: 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 an

(_)  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.