Question

1. a disc jockey has to choose three songs for the last few minutes of his evening show. if there are nine songs that he feels are appropriate for that time slot, then how many ways can he choose and arrange to play in order three of those nine songs? options: ( 9c3 = \frac{9!}{3! (9 - 3)!} ), ( 3p9 = \frac{3!}{9!} ), ( 9p3 = \frac{9!}{(9 - 3)!} ), ( 9p3 = \frac{9!}{3!} )

Explanation:

Step1: Identify the problem type

This is a permutation problem because we are choosing and arranging (order matters) 3 songs from 9. The formula for permutations \( _nP_r \) (permutations of \( n \) items taken \( r \) at a time) is \( _nP_r=\frac{n!}{(n - r)!} \).

Step2: Match with the formula

Here, \( n = 9 \) (total songs) and \( r = 3 \) (songs to choose and arrange). So the correct formula should be \( _9P_3=\frac{9!}{(9 - 3)!} \).

Let's analyze the other options:

• \( _9C_3=\frac{9!}{3!(9 - 3)!} \) is for combinations (order doesn't matter), so this is incorrect.
• \( _3P_9=\frac{3!}{9!} \) is invalid because \( n\) (9) should be greater than or equal to \( r\) (3) and the formula is misapplied.
• \( _9P_3=\frac{9!}{3!} \) is the formula for combinations, not permutations, so it's incorrect.

Answer:

\( 9P3=\frac{9!}{(9 - 3)!} \) (the third option)