# In how many ways can a committee of 1 man and 2 women be constituted out of 2 men and 3 women?

• Last Updated : 22 Dec, 2021

Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

Permutation Formula

In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

nPr = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group

r = subset size, the number of things to be selected from the group

Combination

A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Combination Formula

In combination r things are picked from a set of n things and where the order of picking does not matter.

nsub>r =n!⁄((n-r)! r!)

Here,

n = Number of items in set

r = Number of things picked from the group

Note: In the same example, we have dissimilar instance for permutation and combination. For permutation, AB and BA are two different object but for choosing, AB and BA are same.

### In how many ways can a committee of 1 man and 2 women be constituted out of 2 men and 3 women?

Here we use the formula for choosing r objects from n different objects.

The number of ways of choosing r objects from n different objects is given by nCr and

the value of nCr is given as nCr = n! ⁄ r!(n−r)!

Here sequence doesn’t matter

The formula of selecting 1 man from 2 men: 2C1

= 2! ⁄ 1!(2-1)!

= 2! ⁄ 1!1!

= 2

The formula of selecting 2 women from 3 women: 3C2

= 3! ⁄ 2!(3-2)!

= 3! ⁄ 2!1!

= 3×2! ⁄ 2!

= 3

No of ways of selecting 1 man and 2 women = 2C1 × 3C2

= 2×3

= 6

Total number of Person = 2 Men + 3 Women = 5

Person required in committee = 3

No of ways = 5C3 = 10

### ​Similar Questions

Question 1: A group of 4 people is to be constituted from a group of 3 men and 4 women. In how many ways can this be done? How many of these committees would consist of 2 men and 2 women?

Here we use the formula for choosing r objects from n dissimilar objects.

The number of ways of choosing r objects from n different objects

given by nCr and

the value of nCr is given as n Cr = n! ⁄ r!(n−r)! .

Here sequence doesn’t matter

The formula of selecting 2 man from 3 men: 3C2

= 3! ⁄ 2!(3-2)!

= 3! ⁄ 2!1!

= 3×2!/2!

= 3

The formula of selecting 2 women from 4 women: 4C2

= 4! ⁄ 2!(4-2)!

= 4! ⁄ 2!2!

= 4×3×2! ⁄ 2!×2!

= 4×3/2

= 2×3

= 6

No of ways to make a committee of 3 person = 3C2 × 4C2

= 3×6

= 18ways

Total number of Person = 3 Men + 4 Women = 7

Person required in committee = 4

N

ways = 7C4 = 35

Question 2: A group of 6 people is to be constituted from a group of 4 men and 4 women. In how many ways can this be done? How many of these committees would consist of 3 men and 3 women?

Here we use the formula for choosing r objects from n different objects.

The number of ways of choosing r objects from n different objects is given by nCr and

the value of nCr is given as nCr = n! ⁄ r!(n−r)! .

Here sequence doesn’t matter

The formula of selecting 3 man from 4 men: 4C3

= 4! ⁄ 3!(4-3)!

= 4! ⁄ 3!1!

= 4×3!/3!

= 4

The formula of selecting 3 women from 4 women: 4C3

= 4! ⁄ 3!(4-3)!

= 4! ⁄ 3!1!

= 4×3! ⁄ 3!

= 4

No of ways to make a committee of 6 person = 4C3 × 4C3

= 4×4

= 16 ways

Total number of Person = 4 Men + 4 Women = 8

Person required in committee = 6

No of ways = 8C6 = 28

Question 3: A team of 12 people is to be constituted from a group of 10 men and 8 women. In how many ways can this be done? How many of these committees would consist of 7 men and 5 women?

Here we use the formula for choosing r objects from n different objects.

The number of ways of choosing r objects from n different objects is given by nCr and

the value of nCr is given as nCr = n! ⁄ r!(n−r)! .

Here sequence doesn’t matter

The formula of selecting 7 man from 10 men: 10C7

= 10! ⁄ 7!(10-7)!

= 10! ⁄ 7!3!

= 10×7!/3!

= 120

The formula of selecting 5 women from 8 women: 8C5

= 8! ⁄ 5!(8-5)!

= 8! ⁄ 5!3! =  8×7×6×5!/5!×3×2×1

= 4×7×2

= 56

No of ways to make a committee of 12 person = 10C7 × 8C5

= 120 × 56

= 6,720 ways

Total number of Person = 10 Men + 8 Women = 18

Person required in committee = 12

No of ways = 18C12 = 18,564

Question 4: A group of 9 people is to be constituted from a group of 7 men and 5 women. In how many ways can this be done? How many of these committees would consist of 6 men and 3 women?

Here we use the formula for choosing r objects from n different obj

.

The number of ways of choosing r objects from n different objects is given by nCr and

the value of nCr is given as nCr= n! ⁄ r!(n−r)! .

Here sequence doesn’t matter

The formula of selecting 6 man from 7 men: 7C6

= 7! ⁄ 6!(7-6)!

= 7! ⁄ 6!1!

= 7×6!/6!

= 7

The formula of selecting 3 women from 5 women: 5C3

= 5! ⁄ 3!(5-3)!

= 5! ⁄ 3!2!

= 5×4×3! ⁄ 3!×2×1

= 10

No of ways to make a committee of 6 person  = 7C6 × 5C3

= 7×10

= 70 ways

Total number of Person =  7 Men + 5 Women = 12

Person required in committee = 9

No of ways =  12C9 = 220

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