10 students have volunteered for a committee. 7 of them are seniors and 3 of them are juniors. 1) how many ways are there to select a committee of 8 students? 10!/(8!2) 2) how many ways are there to select a committee with 3 seniors and 5 juniors? 7!/(3!/4!)*(3!/(5!/(2)!)) 3) suppose the committee must have 8 students (either juniors or seniors) and that one of the 8 must be selected as chair. how many ways are there to make the selection?

(c) The number of ways to select a committee having 8 students (either juniors or seniors) and that one of the 8 must be selected as chair = 360 ways

What is combination?

A combination in mathematics is a choice made from a group of separate elements where the order of the selection is irrelevant.

Given that 10 students have volunteered for a committee. 7 of them are seniors and 3 are juniors.

a) Need to find the number of ways of selecting a committee of 8 students

The number of ways of selecting a committee of 8 students = [tex]10_C_8[/tex]

[tex]10C_8= \frac{10!}{8!(10-8)!}= \frac{10!}{8!(2!)}=45[/tex]

The number of ways of selecting a committee of 8 students = 45.

(b) Need to find the number of ways to select a committee with 3 seniors and 2 juniors

The number of ways to select a committee with 3 seniors and 2 juniors

= [tex]7_C_3(3_C_2)[/tex]

[tex]7_C_3*3_C_2 = \frac{7!}{3!*4!} *\frac{3!}{2!*1!} = 35 * 3 = 105[/tex]

The number of ways to select a committee with 3 seniors and 2 juniors are 105.

(c) Need to find the ways to make the select the committee must have 8 students (either juniors or seniors) and that one of the 8 must be selected as chair.

Number of ways = [tex]10_C_8*8_C_1[/tex]

[tex]10_C_8*8_C_1 = \frac{10!}{8!*2!} *\frac{8!}{7!*1!} = 45*8 = 360[/tex]

The number of ways to select a committee having 8 students (either juniors or seniors) and that one of the five must be selected as chair = 360 ways.

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