Final answer:
To determine the number of ways to select a lead and an understudy from 30 people, use the permutation formula P(30,2), which yields 870 different combinations.
Step-by-step explanation:
The question asks for the number of different ways to select a lead and an understudy from a group of 30 people auditioning for a school play. This is a problem of permutations because the order of selection matters - the first person selected will be the lead, and the second will be the understudy.
To find the total number of permutations, we apply the formula for permutations of n items taken r at a time, which is P(n,r) = n! / (n-r)!. In this case, we want the permutations of 30 people taken 2 at a time, so we calculate P(30,2) = 30! / (30-2)! = 30! / 28!.
Calculating this, we get:
30 × 29 = 870 (This is because when simplifying the factorial, all terms from 28! downwards cancel out, leaving only 30 × 29.)
Therefore, there are 870 different ways to select a lead and an understudy for the play.