Evaluate 12C10

Combination is a method to calculate the total number of outcomes where the order of the outcomes does not matter. This implies that the objects can be selected in any order. To summarise, combinations of any two integers denote all the possible re-arrangements of these numbers. For instance, let us suppose we have three names, A,B and C, selecting any two possible names from these, will correspond to AB, BC, AC. In combinations, AB is considered to be equivalent to BA. The total number of combinations can be obtained by the following formula, 

C(n, r) = n!/(n – r)!r!

where, 

n = total number of objects in the set

r = number of choosing objects from the set

A combination is an unordered permutation. 

There are basically two types of combinations: Combination with repetition or without repetition 

	With repetition	Without repetition
Combinations	(r + n − 1)!r!(n − 1)!	n!r!(n − r)!

Evaluate ¹²C₁₀

Solution:

We know, 

C(n, r) = n!/(n – r)!r!

Here, we have, 

n = 12

r = 10

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

We get,

C(12, 10) = 12!/(12 – 10)!10!

On solving, we get, 

C(12, 10) = 12!/(2)!10!

Now, 

C(12, 10) = 12 × 11 × 10!/2! × 10!

C(12, 10) = 12 × 11/ 2

On further simplifying, we get, 

C(12, 10) = 132/2

C(12, 10) = 66

Therefore,

C(12,10). C(n,r) = n!/(n-r)!r! = 66

Sample Questions

Question 1. Find C(6,4). C(n,r) = n!/(n-r)!r!

Solution:

Here,

n = 6

r = 4

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

We get,

C(6, 4) = 6!/(6 – 4)!4!

C(6, 4) = 6!/(2)!4!

C(6, 4) = 6 × 5 × 4!/2! × 4!

C(6, 4) = 6 × 5/ 2

C(6, 4) = 30/2

C(6, 4) = 15

Therefore,

C(6, 4). C(n,r)=n!/(n-r)!r! = 15

Question 2. Evaluate C(24, 20). C(n, r) = n!/(n-r)!r!

Solution:

Here,

n = 24

r = 20

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

We get,

C(24, 20) = 24!/(24 – 20)!20!

C(24, 20) = 24!/(4)!20!

C(24, 20) = 24 × 23 × 22 × 21 × 20!/4! × 20!

C(24, 20) = 24 × 23 × 22 × 21/ 4 × 3 × 2 × 1 

C(24, 20) = 255024/24

C(24, 20) = 10626

Therefore,

C(24, 20). C(n,r) = n!/(n-r)!r! = 10626

Question 3. Find out C(18, 16). C(n, r) = n!/(n-r)!r!

Solution:

Here,

n = 18

r = 16

Putting the values of n and r in C(n, r) = n!/(n – r)!r!

We get,

C(18, 16) = 18!/(18 – 16)!16!

C(18, 16) = 18!/(2)!18!

C(18, 16) = 18 × 17 × 16!/2! × 16!

C(18, 16) = 24 × 23 × 22 × 21/ 2 

C(24, 20) = 255024/24

C(24, 20) = 10626

Therefore,

C(24, 20). C(n,r) = n!/(n-r)!r! = 10626

Question 4. Difference Between Permutation and Combination?

Solution:

Permutation	Combination
Arranging people, digits, numbers, alphabets, letters, and colours	Selection of menu, food, clothes, subjects, team.
Picking a team captain, pitcher and shortstop from a group.	Picking three team members from a group.
Picking two favourite colours, in order, from a colour brochure.	Picking two colours from a colour brochure.
Picking first, second and third place winners.	Picking three winners.

Question 5. Find out in how many a group of 7 boys and 10 girls, can be selected from 9 boys and 12 girls?

Solution:

Here we have to find,

To select 7 boys out of 9 boys = ⁹C₅ ways = 9!/(9-5)!5!

= 9!/4!5!

= 9 × 8 × 7 × 6 × 5!/4 × 3 × 2 × 1 × 5!

= 3024/24

= 126

To select 10 girls out of 12 girls = ¹²C₁₀ ways = 12!/(12 – 10)!10!

= 12 × 11 × 10!/2!10!

= 132/2

= 66

Now, 

The group can be selected in the 66 * 126 ways 

= 8316 ways