Question

Question 1 160 pts Question 1. A class consists of 4 boys and 5 girls. Answer the following questions: (a) In how many different ways can they be arranged along a line such that the boys are all together and the girls are also all together? (b) In how many different ways can the class be arranged along a circle? (c) In how many different ways can a group of 4 students be selected from the class such that there is at least one girl in it? (d) In how many different ways can a group of 4 students be selected from the class such that there is an odd number of girls in it?

Answer 1

a) Since we want all boys to be together and all girls to be together, consider all boys as one group and all girls as one group. Then we have two groups that can be arranged in 2 ways. Boys ahead, girls back or girls ahead, boys back. Howver, each group can be shuffled within themselves. Such that the group of boys can be arranged in 4! ways and the group of girls can be arranged in 5! ways. Hence the total number of required arrangements are 2*4!*5! = 5760.

b) The number of ways arranging n items along a fixed circle is given to be (n-1)!.
Hence, the number of ways of arranging 9 students along a fixed circle is 8! = 40320.

c) We need a group of 4 students with atleast 1 girl in it. Such a group can be chosen by fixing 1 girl member of the group and then selecting 3 students arbitrarily from the remaining 8 students. This can be done in  $\small ^8C_3 = 56$ ways. But since a girl member can be fixed in 5 ways for 5 girls, the total number of ways for forming the group is 56*5 = 280.

d) #{Groups of 4 students with odd no. of girls} = #{Groups of 4 students with exactly 1 girl} + #{Groups of 4 students with exactly 3 girls}
Now, a group of 4 students with exactly 1 girl can be chosen by chosing 1 girl among 5 girls and chosing 3 boys among 4 boys. Total ways to do this is $\small ^5C_1*^4C_3 = 5*4 = 20$ .
Similary, a group of 4 students with exactly 3 girls can be chosen by chosing 3 girls among 5 girls and chosing 1 boy among 4 boys. Total ways to do this is $\small ^5C_3*^4C_1= 10*4 = 40$ .
Hence, total number of required ways are 40+20 = 60.