Selection Problems in Combinations: A Case Study

In this article, we explore a specific selection problem involving the selection of 3 women from a group of 15 women under certain conditions. The conditions are that one particular woman is always included, and two other particular women are always excluded. We will break down the process step-by-step to determine the total number of ways to achieve this selection.

Problem Restatement

The problem statement can be restated as follows: 'In how many ways can 3 women be selected out of 15 women if one particular woman is always included and two particular women are always excluded?'

Step-by-Step Solution

Identifying the Women Involved

We have a total of 15 women. Let us denote them as:

W1 - One particular woman who is always included. W2 - One particular woman who is always excluded. W3 - Another particular woman who is also always excluded.

Adjusting the Total Count

Since W1 is always included, we do not need to consider her in the selection process. Additionally, we need to exclude W2 and W3. Therefore, the pool for selection is reduced by 3 women out of the total 15.

Counting the Remaining Women

The adjusted number of women we need to select from is:

15 - 1 - 2 12 women

Selecting the Remaining Women

From the remaining 12 women, we need to select 2 more women to form a group of 3. The number of ways to choose 2 women from 12 can be calculated using the combination formula:

(binom{n}{r} frac{n!}{r!(n-r)!})

Calculating the Combinations

Using the combination formula with n12 and r2:

(binom{12}{2} frac{12!}{2!(12-2)!} frac{12!}{2!10!} frac{12 times 11}{2 times 1} 66)

Hence, there are 66 ways to select 3 women under the given conditions.

Alternative Approach

Let#39;s consider another approach to solving the problem:

One Woman is Always Included

This means there is one way to select this particular woman (W1) from the three women.

Two Women are Always Excluded

With these two women excluded, we are left with:

15 - 1 - 2 12 women to choose from for the second and third selections.

The second woman can be selected in 12 ways, and the third woman can be chosen in 11 ways from the remaining 11 women.

The total number of ways to select the three women in this manner is:

12 (options for the second woman) x 11 (options for the third woman) 132 ways.

However, it is important to note that the order of selection does not matter in this context, and thus this approach overcounts the combinations. Therefore, the primary and correct approach through combinations is more accurate, yielding 66 ways.

Case Study Example

Suppose we have a specific woman, say A, who is always included. Then we need to exclude two particular women, let#39;s say B and C. After excluding B and C, we are left with:

15 - 1 - 2 12 women

Out of these 12 women, we need to select 2 more to make up a group of 3. This can be done in:

(binom{12}{2} frac{12!}{2!(12-2)!} frac{12 times 11}{2 times 1} 66)

Therefore, there are 66 ways to choose 3 women from 15 women such that one particular woman is always included and two particular women are always excluded.

Conclusion

In conclusion, we have demonstrated two methods to solve the selection problem. Both methods confirm that there are 66 ways to select 3 women from 15 women under the conditions that one particular woman is always included and two particular women are always excluded.