After solving by **combination **all these **individually A)** 21 , 15 , 315 as a result respectively for three parts **B)**715 methods as a result. **C**)210 alternative four-person committees

What distinguishes the two Mathematical terms like permutation and combination ?

Permutation and combination are used to** count the number** of possible arrangements or choices for an object. Permutation and combination are **primarily distinguished** by the fact that permutation is used when order counts and combination is used when order is irrelevant. ¹³.

**For instance, **if you have the three distinct objects A, B, and C, there are only three potential combinations (ABC, AB, AC) and six possible permutations of these objects (ABC, ACB, BAC, BCA, CAB, CBA).

Let's tackle each issue individually.

(a) The **combinations formula**, nCr = n! / (r! * (n-r)!, can be used to determine how many ways there are to select two women from the group of seven women for the committee. n represents the total number of items, r represents the number of items to be selected, and! denotes factorial (product of all positive integers up to that number). nCr = 7C2 = 7! / (2! * (7-2)!) = 21 ways as a result.

The same method may be used to determine how many **ways** there are to select two men from a group of six men for the committee: nCr = n! / (r! * (n-r)!, where n is the total number of items, r is the number of items to select,! means factorial, and (product of all positive integers up to that number). Since nCr = 6C2 = 6! / (2! * (6-2)!) = 15 ways ,this means that.

The number of ways to select two women by the number of ways to select two men can be multiplied to find the total number of alternative four-person committees that can be formed if the committee must include two women and two men. Hence, there are 315 different four-person committees that might exist (21 * 15) .

(b) Using the combinations formula, nCr = n! / (r! * (n-r)!, where n is the total number of things, r is the number of items to choose, and! signifies factorial, one can determine the total number of different four-person committees that may be formed.(product of every positive integer up to that amount). nCr = (7+6)C4 = 13C4 = 13! / (4! * (13-4)!) = 715 methods as a result.

(c)** Multiplying the number **of ways to select three women by the number of ways to select one man yields the total number of alternative four-person committees that can be formed if the committee must be made up of three women and one male. Thus, there are 210 alternative four-person committees that might exist if 7C3 * 6C1 = 35 * 6 = 210.

By using the combinations formula, nCr = n! / (r! * (n-r)!, where n is the total number of items, r is the number of items to choose from, and! means factorial (product of all items), it is possible to determine the total number of different four-person committees that can be formed if the committee must consist of **four women** and zero men.

nCr = 7C4 thus equals 7! / (4! * (7-4)!) = 35 ways 3.

The following formula can be used to determine how many alternative four-person committees are conceivable if a majority of the members must be women:

- There must be one guy on a committee if there are three women. By combining the number of ways to select three women with the number of ways to **select one man**, which equals 7C3 * 6C1 = 210, it is possible to determine the total number of such committees.

- A committee has no men when there are four women on it.

. Combinations formula nCr = n! / (r! * (n-r)!, where n is the total number of items, r is the number of things to choose from, and! means **factorial**, can be used to get the total number of such committees (product of all positive integers up to that number). Because of this, nCr = 7C4 = 35.

As a result, 210 + 35 = 245 distinct** four-person committees** are feasible if a majority of them must be women .

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