C(n,12) = C(n,8) N = ?a)12b)10c)20d)14Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Answer

C (n,r) = C (n,n-r)
from question
r=8 , n-r=20
by adding
n=20

Answer

To solve this problem, we need to use the concept of combinations (denoted by C). The formula for combinations is:

C(n, r) = n! / (r! * (n-r)!)

where n is the total number of items and r is the number of items being chosen.

In this problem, we are given:

C(n, 12) = C(n, 8)

We need to find the value of n.

Simplifying the equation using the combination formula, we have:

n! / (12! * (n-12)!) = n! / (8! * (n-8)!)

Canceling out the common terms on both sides, we get:

1 / (12! * (n-12)!) = 1 / (8! * (n-8)!)

Now, we can cancel out the factorial terms as well:

1 / (n-12)! = 1 / (n-8)!

Since the denominators on both sides are equal, the numerators must also be equal:

1 = (n-12)! / (n-8)!

We can simplify this further by multiplying both sides by (n-8)!:

(n-8)! = (n-12)!

Now, we can see that (n-8)! = (n-12)!, which means that n-8 = n-12.

Solving this equation, we get:

-8 = -12

This equation is not possible, as -8 is not equal to -12.

Therefore, there is no value of n that satisfies the given equation.

The correct answer is that there is no valid value for n.

Answer

nC12 = nC8
n/ 12 (n - 12) = n/8 (n - 8)
12 = n - 8
n = 12 + 8 = 20
n = 20.

Similar JEE Doubts

If n(

Pairs of consecutive even positive integers, both of which are larger than 7 such that their sum is less than 28, are

If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12} and C = {4, 5, 6, 12, 14}, then(A ∪ B) ∩ (A ∪ C) is equal to

If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12}, C = {4, 5, 6, 12, 14} then (A∩B)∪(A∩C) is equal to

The co-efficient of x4in the expansion of (1 –x + 2x2)12is

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Mathematics (Maths) for JEE Main & Advanced

Chemistry for JEE Main & Advanced

Study Plan for JEE

Top Courses for JEE

Top Courses for JEE

Related Content

JEE Courses

Mock Test Series

Past Year Papers

Additional Courses

Company