Question

find the value of n in ⁿc₂=28.​

Answer 1

Step-by-step explanation:

Given :-

ⁿc₂=28.

To find:-

Find the value of n ?

Solution:-

Given that

ⁿc₂=28.

We know that ⁿcr = n!/[r!×(n-r)!]

We have r = 2

Now ,

=> ⁿc₂ = n!/[2!×(n-2)!]

=> n!/[2×(n-2)!] = 28

=> [n×(n-1)×(n-2)!]/[2×(n-2)!] = 28

On cancelling (n-2)! then

=> (n)(n-1)/2 =28

=>n(n-1) = 28×2

=> n(n-1) = 56

=>n^2-n = 56

=>n^2-n-56=0

=> n^2-8n+7n-56 = 0

=> n(n-8)+7(n-8) = 0

=> (n-8)(n+7) = 0

=> n-8 = 0 or n+7 = 0

=> n=8 or n= -7

But n can not be negative

Therefore, n = 8

Answer:-

The value of n for the given problem is 8

Check:-

If n= 8, then ⁿc₂

=> 8c2

=>8!/(2!×6!)

=>8×7×6!/(2×6!)

=> 8×7/2

=> 56/2

=> 28

Verified the given relation.

Used formula:-

• ⁿcr = n!/[r!×(n-r)!]

Answer 2

n = 8

Refer the attachments... ✔️