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Question: If \({}^{2n}{C_3}:{}^n{C_2} = 44:3\) then \(n\)= \(A) 6\) \(B) 7\) \(C) 8\) \(D) 9\)...

If 2nC3:nC2=44:3{}^{2n}{C_3}:{}^n{C_2} = 44:3 then nn=
A)6A) 6
B)7B) 7
C)8C) 8
D)9D) 9

Explanation

Solution

Here we have to find the value of nn.
It is given the combination to solve it using the combination formula and we split some terms and finally we get the required answer.

Formula used: That is nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} firstly then simplifying the equation and get the required result.

Complete step-by-step solution:
It is given that the question stated as n∈Nn \in \mathbb{N} such that 2nC3:nC2=44:3{}^{2n}{C_3}:{}^n{C_2} = 44:3
Here we have to find out the value for nn
Using the formula ofnCr{}^n{C_r}, nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
We can write the given ratio in an expression as:
⇒2n!3!(2n−3)!:n!2!(n−2)!=44:3\Rightarrow \dfrac{{2n!}}{{3!(2n - 3)!}}:\dfrac{{n!}}{{2!(n - 2)!}} = 44:3
Or we can write the term on equivalently as a fraction:
We just divided the one fraction to the other one
⇒2n!3!(2n−3)!n!2!(n−2)!=443\Rightarrow \dfrac{{\dfrac{{2n!}}{{3!(2n - 3)!}}}}{{\dfrac{{n!}}{{2!(n - 2)!}}}} = \dfrac{{44}}{3}
Now we just reversed both the fractions
⇒2n!3!(2n−3)!×2!(n−2)!n!=443\Rightarrow \dfrac{{2n!}}{{3!(2n - 3)!}} \times \dfrac{{2!(n - 2)!}}{{n!}} = \dfrac{{44}}{3}
And if we expand the above expression as follows:
⇒(2n)(2n−1)(2n−2)(2n−3)......13!(2n−3)!×2!(n−2)!n(n−1)(n−2)....1=443\Rightarrow \dfrac{{(2n)(2n - 1)(2n - 2)(2n - 3)......1}}{{3!(2n - 3)!}} \times \dfrac{{2!(n - 2)!}}{{n(n - 1)(n - 2)....1}} = \dfrac{{44}}{3}
Cancelling the same terms from the numerators and denominators we get:
⇒(2n)(2n−1)2(n−1)3!×2!n(n−1)=443\Rightarrow \dfrac{{(2n)(2n - 1)2(n - 1)}}{{3!}} \times \dfrac{{2!}}{{n(n - 1)}} = \dfrac{{44}}{3}
Expanding the factorial term and we get
⇒2(n)(2n−1)2(n−1)6×2n(n−1)=443\Rightarrow \dfrac{{2(n)(2n - 1)2(n - 1)}}{6} \times \dfrac{2}{{n(n - 1)}} = \dfrac{{44}}{3}
Again we cancel the same terms from the numerators and denominators, we get
⇒4(2n−1)3=443\Rightarrow \dfrac{{4(2n - 1)}}{3} = \dfrac{{44}}{3}
We can cancel the multiple of terms on the opposite sides like 44by444by4 and 3by33by3
Hence we found an equation
⇒2n−1=11\Rightarrow 2n - 1 = 11
On adding −1 - 1 on both side, we can have
⇒2n=11+1\Rightarrow 2n = 11 + 1
By applying adding operation in the right hand side
⇒2n=12\Rightarrow 2n = 12
On dividing 22 on both side we get,
⇒n=122\Rightarrow n = \dfrac{{12}}{2}
We get,
∴n=6\therefore n = 6

Hence the correct option is (A)(A) that is 66

Note: In this problem we have to verify the answer whether we found the value is correct or not
Verifying the answer:
It is given that 2nC3:nC2=44:3{}^{2n}{C_3}:{}^n{C_2} = 44:3
4(2n−1)3=443\dfrac{{4(2n - 1)}}{3} = \dfrac{{44}}{3}
We just put the value of n=6n = 6 we get
4(2×6−1)3=443\dfrac{{4(2 \times 6 - 1)}}{3} = \dfrac{{44}}{3}
On multiply the terms we get,
4(12−1)3=443\dfrac{{4(12 - 1)}}{3} = \dfrac{{44}}{3}
On subtracting the terms,
4(11)3=443\dfrac{{4(11)}}{3} = \dfrac{{44}}{3}
Let us multiply the numerator we get
443=443\dfrac{{44}}{3} = \dfrac{{44}}{3}
Hence the left hand side is equal to the right hand side.