Question

answer the questions below.

(a) an experiment involves 32 participants. from these, a group of 5 participants is to be tested under a special condition. how many groups of 5 participants are possible?

(b) how many different committees of size 4 can be formed from 13 people?

Explanation:

Part (a)

Step1: Identify the problem type

This is a combination problem since the order of participants in the group does not matter. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 32 \) (total participants) and \( k = 5 \) (participants in the group).

Step2: Calculate the combination

First, calculate the factorials:

• \( 32! = 32\times31\times30\times29\times28\times27! \)
• \( 5! = 5\times4\times3\times2\times1 = 120 \)
• \( (32 - 5)! = 27! \)

Substitute into the combination formula:
\( C(32, 5)=\frac{32!}{5!(32 - 5)!}=\frac{32\times31\times30\times29\times28\times27!}{120\times27!} \)

Cancel out the \( 27! \) terms:
\( C(32, 5)=\frac{32\times31\times30\times29\times28}{120} \)

Calculate the numerator: \( 32\times31 = 992 \), \( 992\times30 = 29760 \), \( 29760\times29 = 863040 \), \( 863040\times28 = 24165120 \)

Then divide by 120: \( \frac{24165120}{120}=201376 \)

Step1: Identify the problem type

This is also a combination problem. The formula for combinations is \( C(n, k)=\frac{n!}{k!(n - k)!} \), where \( n = 13 \) (total people) and \( k = 4 \) (people in the committee).

Step2: Calculate the combination

First, calculate the factorials:

• \( 13! = 13\times12\times11\times10\times9! \)
• \( 4! = 4\times3\times2\times1 = 24 \)
• \( (13 - 4)! = 9! \)

Substitute into the combination formula:
\( C(13, 4)=\frac{13!}{4!(13 - 4)!}=\frac{13\times12\times11\times10\times9!}{24\times9!} \)

Cancel out the \( 9! \) terms:
\( C(13, 4)=\frac{13\times12\times11\times10}{24} \)

Calculate the numerator: \( 13\times12 = 156 \), \( 156\times11 = 1716 \), \( 1716\times10 = 17160 \)

Then divide by 24: \( \frac{17160}{24}=715 \)

Answer:

(a):
\( 201376 \)

Part (b)