Finding the Number of Ways to Select Members with Exclusions

This question asks us to find the number of ways to select a group of 5 members from a larger group of 10 members, with a specific condition that two particular members must not be included in the selected group.

This is a problem that can be solved using the concept of combinations, as the order in which the members are selected does not matter.

Analyzing the Problem Conditions

• Total number of members available initially: 10
• Number of members to be selected for the final group: 5
• Condition: Two particular members must always be excluded from the selection.

Adjusting the Selection Pool

Since two specific members are always excluded, we first need to remove these two members from the total group of 10. The remaining members form the pool from which our selection of 5 must be made.

Number of members remaining after exclusion = Total members - Members to be excluded

Number of members remaining = $10 - 2 = 8$

Now, the problem is simplified to selecting 5 members from this reduced pool of 8 members.

Applying the Combination Formula

The number of ways to select $k$ items from a set of $n$ items, without regard to the order of selection, is given by the combination formula:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

In our case:

• $n$ = Number of members available for selection = 8
• $k$ = Number of members to be selected = 5

So, we need to calculate $\binom{8}{5}$.

Calculating the Number of Combinations

Let's calculate $\binom{8}{5}$ using the formula:

$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}$

Expanding the factorials:

$\binom{8}{5} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)(3 \times 2 \times 1)}$

We can cancel out the $5!$ term from the numerator and denominator:

$\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

Calculating the denominator: $3 \times 2 \times 1 = 6$.

So, $\binom{8}{5} = \frac{8 \times 7 \times 6}{6}$

Cancel out the 6 from numerator and denominator:

$\binom{8}{5} = 8 \times 7$

$\binom{8}{5} = 56$

Therefore, there are 56 ways to select 5 members out of 10, such that the two particular members must always be excluded.

Summary of Calculation

The number of ways to select 5 members from the 8 available members is 56.