Question

if nC3=20 then find the value of n​

Answer 1

If nC3 = 20, then the value of "n" will be 6

Given :

nC3 = 20

To Find :

Value of n

Solution :

According to permutations and combinations;

$nCr = \frac{n!}{r!(n-r)!}$

$n! = n *(n-1)*(n-2)*...............*3*2*1$

From above formula we can also write n! as follows

$n! = n*(n-1)!$  or  $n! = n*(n-1)*(n-2)!$ depending on the requirement

Now,

using the above relation let us rewrite our expression given in the question;

$nC3 = 20$

⇒ $nC3 = \frac{n!}{3! * (n-3)!}$

⇒ $nC3 = \frac{n*(n-1)*(n-2)*(n-3)!}{3! * (n-3)!}$

Let us cancel out $(n-3)!$ terms in both numerator and denominator and replace $nC3$ with the value given in the question;

⇒ $20 = \frac{n * (n-1) * (n-2)}{3!}$

⇒ $20 = \frac{n * (n-1) * (n-2)}{3*2*1}$

⇒ $20 = \frac{n * (n-1) * (n-2)}{6}$

By cross multiplication;

⇒ (20 * 6) = n * (n-1) * (n-2)

⇒ 120 = n * (n² - 3n + 2)

⇒ 120 = n³ - 3n² + 2n

Let us rearrange the above expression into a cubic equation and solve for n value;

⇒ n³ - 3n² + 2n - 120 = 0

⇒ (n - 6) (n - (-1.5 + 4.2i)) (n - (-1.5 - 4.2i )) = 0

⇒ n = 6 or n = -1.5 - 4.2i or n = -1.5 + 4.2i

But, only n = 6 is a valid value because the values of n, r in permutations and combinations  should be real numbers only.

Here, out of all three values for n, only one value is real number.

Hence, when nC3 = 20, value of n will be 6.

Answer 2

Value of n = 6

Given:

ⁿc₃ = 20

To Find:

Value of n

Solution:

According to permutations and combinations

ⁿ$C_{r}$ = $\frac{n!}{r!(n-r)!}$

here,

r = 3

ⁿ$C_{r}$ = 20

n! = n * (n-1) * (n-2) * ... * 3 *2*1

Substituting this in our equation,

=> ⁿ$C_{r}$ = $\frac{n * (n-1)*(n-2)*(n-3)!}{(3!)(n-3)!}$

=> 20 =  $\frac{n * (n-1)*(n-2)*(n-3)!}{(3!)(n-3)!}$

We can cancel (n-3)! from numerator and denominator as it is common

=> 20 =  $\frac{n * (n-1)*(n-2)}{(3!)}$

=> 20 = $\frac{n * (n-1)*(n-2)}{3*2*1}$

=> 20 = $\frac{n * (n-1)*(n-2)}{6}$

=> 20 *6 = n*(n-1)*(n-2)

=> 120 = $n^{3} - 3n^{2} + 2n$

We can rearrange the above expression into a cubic equation and solve for n,

=> $n^{3} - 3n^{2} + 2n$ - 120 = 0

=> (n - 6) (n - (-1.5 + 4.2i)) (n - (-1.5 - 4.2i )) = 0

⇒ n = 6 or n = -1.5 - 4.2i or n = -1.5 + 4.2i

Only 6 is possible as the others are imaginary roots

n must be a real number only

Hence, when ⁿc₃ = 20, the value of n will be 6.

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