Question

2np3=2*nP4 Find the value of n

Answer 1

Answer:

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

2(2n-1)=n^2-5n+6

n^2-9n+8=0

(n-1)(n-8)=0

n=8

Answer 2

Value of n is 8.

Given:

• $^{2n} P_3 = 2 \times \: ^{n}P_4$

To find:

• Find the value of n.

Solution:

Formula to be used:

$\bf ^{n} P_r = \frac{ n!}{(n - r)!}$

Step 1:

Open formula of permutation both sides.

$\frac{(2n)!}{(2n - 3)!} = 2 \times \frac{n!}{(n - 4)!}$

or

$\frac{2n(2n - 1)(2n - 2)(2n - 3)!}{(2n - 3)!} = 2 \times \frac{n(n - 1)(n - 2)(n - 3)(n - 4)!}{(n - 4)!}$

or

$\frac{4n(2n - 1) \cancel{(n - 1)}\cancel{(2n - 3)!}}{\cancel{(2n - 3)!}} = 2 \times \frac{n\cancel{(n - 1)}(n - 2)(n - 3)\cancel{(n - 4)!}}{\cancel{(n - 4)!}}$

or

Similarly cut common terms.

$2(2n - 1) = (n - 2)(n - 3)$

Step 2:

Solve for n.

Simplify the equation.

$4 {n} - 2 = {n}^{2} - 3n - 2n + 6$

or

${n}^{2} - 9n + 8 = 0$

or

${n}^{2} - 8n - n + 8 = 0$

or

$n(n - 8) - 1(n - 8) = 0$

or

$(n - 8)(n - 1) = 0$

or

$\bf n = 8$

or

$\bf n = 1$

we have to discard n=1, because in permutation n > r, as we have to choose less from more items.

Thus,

Value of n is 8.

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