Question 1194074
Four girls and three boys are seated randomly in a row. Calculate the following probabilities.
(a) The girls occupy the first three places.
(b) They are arranged to alternate girl, boy, girl, etc.
Answers
(a) 4/35
(b) 1/35
Can someone explain the answers? Thank you.
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<pre>
(a)  The total number of all possible different permutations/arrangements, without any constraints, is

         (4+3)! = 7! = 1*2*3*4*5*6*7.


     If the girls occupy first three position, then the number of all such permutations/arrangements is

         (4*3*2) * (7-3)! = (4*3*2) * (4*3*2*1).


     The probability under question (a) is the ratio of these numbers

         P = {{({(4*3*2)*(4*3*2*1))/(1*2*3*4*5*6*7)}}} = {{{4/35}}}.    <U>ANSWER</U>




(b)  There is only one placing scheme, satisfying conditions (b). It is  (G B G B G B G).


     Inside this scheme, 4! * 3! arrangements are possible.


     In all, there are 7! arrangements of 4 girls and 3 boys.


     The ratio of these two numbers is the sough probability  P = {{{(4!*3!)/7!}}}.


     Reduce this fraction and get the answer  {{{1/35}}}.
</pre>

Solved.