Question

A teacher wants to take 10 students to a park. He can take exactly 3 students at a time and will not take the same group of 3 students more than once. Find the maximum number of times (i) each student can go to the park (ii) the teacher can go to the park.

Answer 1

The number of ways of selecting r things out of n things is $nC_{r}$.

Selecting 3 students out of 10 students (non repeating the same group) is $10C_{3}$.

Note that the teacher will accompany the students every time.

So, the teacher can go to the part $10C_{3}$ times.

Now, $10C_{3}= \frac{10!}{(10-3)!3!}$

$}= \frac{10!}{7!3!}$

$}= \frac{10(9)(8)7!}{7!3!}$

$}= \frac{10(9)(8)}{3(2)(1)}$

= 10 × 3 × 4

= 120 times

Hence, the teacher can go to the park 120 times and this solves (ii).

Now, the number of times each student can go to the park = total number of trips - number of times other students go to the park

Note that the total number of trips is 120 since for each trip, the teacher goes.

Now, the number of trips other students can go is the number of ways of selecting 3 students from the remaining 9 students.

This can be done in $9C_{3}$ ways.

$9C_{3} =\frac{9!}{(9-3)!3!}$

$=\frac{9!}{6!3!}$

$=\frac{9(8)(7)6!}{6!3!}$

$=\frac{9(8)(7)}{3(2)(1)}$

= 3 × 4 × 7

= 84 ways

Hence, the number of times each student can go to the park is 120 - 84 = 36 and this solves (i).