In how many ways a committee of 3 can be made from a total of 10 members?

The 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. This is one basic difference between both the definitions and that is what makes their formulae different from each other.

Permutation

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.

 ⁿP_r = (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, the 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.

ⁿC_r = n! ⁄ ((n – r)! r!)

Here,

n = Number of items in set

r = Number of things picked from the group

In how many ways a committee of 3 can be made from a total of 10 members?

Solution:

In the first part, use the combination formula since order doesn’t matter. 

Then,

n!/r!(n – r)! = 10!/3!(10 – 3)!

= 120.

In the second part the order is required to choose the members from the committee 

For that, use the permutation formula,

n!/(n – r)! = 10!(10 – 3)! 

= 720.

Similar Problems

Question 1: There are 8 men and 10 women in total. How many ways are there to choose 5 men and 6 women to form a committee? 

Solution:

Number of ways

= ⁸C₅ × ¹⁰C₆

= ⁸C₃ × ¹⁰C₄ [∵ⁿC_r = ⁿC_{(n – r)}]

= [(8 × 7 × 6)/(3 × 2 × 1)] × [(10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)]

= 56 × 210

= 11760

Question 2: There are 2 white-colored balls, 3 black-colored balls, 4 red-colored balls. Find the number of ways of drawing 3 balls such that at least 1 black ball is present in the draw from the bag.  

Solution:

There are 2 white coloured balls, 3 black coloured balls, 4 red coloured balls in the bag from which 3 balls need to be drawn in such a way that at least 1 black ball is present there in the draw from the bag. There are 3 possibility.

• Possibility 1: Choosing 3 black balls.
• Possibility 2: Choosing 2 black balls and one non-black ball. 
• Possibility 3: Choosing 1 black ball and 2 non-black balls.  

Number of ways of selecting 3 black balls,

= ³C₃

•  Number of ways to select 2 black balls and 1 non-black ball =  ³C₂ × ⁶C₁
• Number of ways to select 1 black balls and 2 non-black ball =  ³C₁ × ⁶C₂

Total number of ways,

= ³C₃ + ³C₂ × ⁶C₁ + ³C₁ × ⁶C₂

= ³C₃ + ³C₁ × ⁶C₁ + ³C₁ × ⁶C₂ [∵ⁿC_r =  ⁿC_{(n – r)}]

= 1 + (3 × 6) + [3 × (6 × 5)/(2 × 1)]

= 1 + 18 + 45

= 64