In how many ways can a table tennis team of 4 be selected from 9 people if two brothers must be selected…

Question

In how many ways can a table tennis team of 4 be selected from 9 people if two brothers must be selected together or not at all, and two quarreled people cannot be selected together?

ScanSolve AI Tutor · Accepted Answer

Here's how to solve this problem step-by-step:

Total number of people available = 9.
Size of the table tennis team to be selected = 4.

Let the two brothers be A and B.
Let the two quarreled people be X and Y.
The remaining $9 - 4 = 5$ people are others.

We have two main conditions:

Brothers (A, B) must either both be selected or both be left out.
Quarreled people (X, Y) cannot be selected together.

We will consider two cases based on the first condition.

Case 1: Both brothers (A and B) are selected.

Select A and B.
This fills 2 spots on the team.
Number of remaining spots to fill = $4 - 2 = 2$ .
Number of remaining people to choose from = $9 - 2 = 7$ (these include X, Y, and the 5 others).
Select the remaining 2 people from the 7 available, ensuring X and Y are not together.
First, calculate the total ways to choose 2 people from 7 without any restriction:
$C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21$
Next, calculate the ways where X and Y are chosen together (which is not allowed). If X and Y are chosen, there is only 1 way to select them both:
$C(2, 2) = 1$
Subtract the disallowed combinations from the total:
Number of ways for Case 1 = $21 - 1 = 20$ .

Case 2: Neither brother (A nor B) is selected.
A and B are not selected.
This means we need to select all 4 team members from the remaining $9 - 2 = 7$ people (these include X, Y, and the 5 others).
Select 4 people from these 7, ensuring X and Y are not together.
First, calculate the total ways to choose 4 people from 7 without any restriction:
$C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$
Next, calculate the ways where X and Y are chosen together. If X and Y are chosen, 2 spots are filled. We need to choose $4 - 2 = 2$ more people from the remaining $7 - 2 = 5$ people (the 5 others).
Number of ways to choose 2 from 5 = $C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$ .
Subtract the disallowed combinations from the total:
Number of ways for Case 2 = $35 - 10 = 25$ .
Sum the possibilities from both cases.
Total number of possible selections = Ways from Case 1 + Ways from Case 2
Total = $20 + 25 = 45$ .

The total number of selections possible is $\text{45}$ .

In how many ways can a table tennis team of 4 be selected from 9 people if two brothers must be selected together or not at all, and two quarreled people cannot be selected together?

Answer

Step-by-step solution

Select A and B.

Select the remaining 2 people from the 7 available, ensuring X and Y are not together.

A and B are not selected.

Select 4 people from these 7, ensuring X and Y are not together.

Sum the possibilities from both cases.

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