Question

a camp counselor and eight campers are to be seated along a picnic bench. in how many ways can this be done if the counselor must be seated in the seventh seat and a camper who has a tendency to engage in food fights must sit to the counselor’s immediate right?
there are \\(\square\\) way(s) this can be done

Explanation:

Step1: Fix the counselor's position

Since the counselor must be in the seventh seat, we fix that position.

Step2: Fix the trouble - maker camper's position

The trouble - maker camper must be to the counselor's immediate right (eighth seat). So, these two positions are fixed.

Step3: Arrange the remaining campers

We have \(7\) remaining campers to arrange in the first \(6\) seats. The number of permutations of \(n\) distinct objects is \(n!\). Here \(n = 7\) (because after fixing two positions, we are arranging \(7\) campers in the remaining \(6\) non - fixed seats. Wait, no, total number of people to arrange: counselor and \(8\) campers. After fixing counselor (7th seat) and the trouble - maker (8th seat), we have \(7\) campers left. The number of ways to arrange \(n\) distinct objects is \(n!\). The formula for permutations of \(n\) distinct objects \(P(n,n)=\frac{n!}{(n - n)!}=n!\). Here \(n=7\) (the remaining \(7\) campers). The number of arrangements of \(n\) distinct objects is \(n!\). So the number of ways \(=7!\)

$$7! = 7\times6\times5\times4\times3\times2\times1=5040$$

Answer:

\(5040\)