Answer to Question #106856 in Statistics and Probability for Piyush

Question

A delegation of 4 students is selected each year from a college to attend the National
Student Association annual meeting. (i) In how many ways can the delegation be
chosen if there are 12 eligible students? (ii) In how many ways if two of the eligible
students will not attend the meeting together? (iii) In how many ways if two of the
eligible students are married and will only attend the meeting together?

Answer 1

(i) In how many ways can the delegation be chosen if there are 12 eligible students?

$$\binom{12}{4}={12! \over 4!(12-4)!}={12(11)(10)(9) \over 1(2)(3)(4)}=495$$

(ii) In how many ways if two of the eligible students will not attend the meeting together?

$$\binom{12}{4}-\binom{12-2}{4-2}=495-{10! \over 2!(10-2)!}=495-45=450$$

(iii) In how many ways if two of the eligible students are married and will only attend the meeting together?

$$\binom{12-2}{4-2}+\binom{12-2}{4}={10! \over 2!(10-2)!}+{10! \over 4!(10-4)!}=$$

$$=45+{10(9)(8)(7) \over 1(2)(3)(4)}=45+210=255$$