10.
Find n if:
(2n)!
n!
7!(2n–7)! 4!(n-4)1 = 24:1
Answers
Hi,
Answer: n = 5
Step-by-step explanation:
Given data:
[(2n)!/7!(2n-7)!] / [(n)!/4!(n-4)!] = 24/1 ……. (i)
Eq. (i) can also be written as,
⇒ ²ⁿC₇ / ⁿC₄ = 24/1
To find: find the value of n
Solution:
Step 1:
Taking the numerator of the L.H.S. first from the eq. (i),
[(2n)!/7!(2n-7)!]
= [(2n)(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)(2n-6)] / 7!
= [(2n)(2n-2)( 2n-4)(2n-6)]*[(2n-1) (2n-3)(2n-5)] / [7*6*5*4*3*2*1]
= [(2* 2 * 2 * 2) * (n)(n-1)(n-2)(n-3)] * [(2n-1)(2n-3)(2n-5)] / 5040
= [16 * n(n-1)(n-2)(n-3)(2n-1)(2n-3)(2n-5)] /5040
= [n(n-1)(n-2)(n-3)(2n-1)(2n-3)(2n-5)] / 315 …… (ii)
Also, taking the denominator of the L.H.S. from the eq. (i),
[(n)!/4!(n-4)!]
= [n(n-1)(n-2)(n-3)] / [4*3*2*1]
= [n(n-1)(n-2)(n-3)] /24 ……. (iii)
Step 2:
From (i), (ii) & (iii), we get
[{n(n-1)(n-2)(n-3)(2n-1)(2n-3)(2n-5)}/315] / [{n(n-1)(n-2)(n-3)}/24] = 24/1
on cancelling the similar terms and rearranging it, we get
⇒ (2n-1)(2n-3)(2n-5) = 315 ….. (iv)
Step 3:
Now, using the prime factorising method we will find the factors of 315 are as follows,
315 = 3 * 3 * 5 * 7 = 9 * 5 * 7 ….. (v)
Comparing eq. (iv) & (v), we get
2n - 1 = 9
⇒ 2n = 9+1
⇒ n = 10/2 = 5
Thus, the value of n is 5.
Let’s verify the value of n that we have calculated by putting it in eq. (i),
L.H.S
[(2*5)!/7!(2*5-7)!] / [(5)!/4!(5-4)!]
= [10!/(7!*3!)] / [5!/(4!*1!)]
= 24/1
= R.H.S.
Hope this is helpful!!!