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Permutations and Combinations question

2023 · 31 Jan · Shift 2 · Q41
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  5. /2023 · 31 Jan · Shift 2 · Q41

Permutations and Combinations question

2023 · 31 Jan · Shift 2 · Q41

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
If 2n+1Pn−1:2n−1Pn=11:21{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_{n}=11: 212n+1Pn−1​:2n−1Pn​=11:21, then n2+n+15n^{2}+n+15n2+n+15 is equal to :
Numerical answer
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Correct answer: 45

  1. We use the permutation formula aPb=a!(a−b)!.^aP_b=\frac{a!}{(a-b)!}.aPb​=(a−b)!a!​.

  2. Compute each term:

2n+1Pn−1=(2n+1)!(2n+1−(n−1))!=(2n+1)!(n+2)!.^{2n+1}P_{n-1}=\frac{(2n+1)!}{(2n+1-(n-1))!}=\frac{(2n+1)!}{(n+2)!}.2n+1Pn−1​=(2n+1−(n−1))!(2n+1)!​=(n+2)!(2n+1)!​.

2n−1Pn=(2n−1)!(2n−1−n)!=(2n−1)!(n−1)!.^{2n-1}P_n=\frac{(2n-1)!}{(2n-1-n)!}=\frac{(2n-1)!}{(n-1)!}.2n−1Pn​=(2n−1−n)!(2n−1)!​=(n−1)!(2n−1)!​.

  1. Given 2n+1Pn−1:2n−1Pn=11:21,^{2n+1}P_{n-1}:{}^{2n-1}P_n=11:21,2n+1Pn−1​:2n−1Pn​=11:21, so (2n+1)!(n+2)!(2n−1)!(n−1)!=1121.\frac{\dfrac{(2n+1)!}{(n+2)!}}{\dfrac{(2n-1)!}{(n-1)!}}=\frac{11}{21}.(n−1)!(2n−1)!​(n+2)!(2n+1)!​​=2111​.

  2. Simplify: (2n+1)!(2n−1)!⋅(n−1)!(n+2)!=1121.\frac{(2n+1)!}{(2n-1)!}\cdot\frac{(n-1)!}{(n+2)!}=\frac{11}{21}.(2n−1)!(2n+1)!​⋅(n+2)!(n−1)!​=2111​.

Now, (2n+1)!(2n−1)!=(2n+1)(2n),\frac{(2n+1)!}{(2n-1)!}=(2n+1)(2n),(2n−1)!(2n+1)!​=(2n+1)(2n), and (n−1)!(n+2)!=1(n+2)(n+1)n.\frac{(n-1)!}{(n+2)!}=\frac{1}{(n+2)(n+1)n}.(n+2)!(n−1)!​=(n+2)(n+1)n1​.

Hence, (2n+1)(2n)(n+2)(n+1)n=1121.\frac{(2n+1)(2n)}{(n+2)(n+1)n}=\frac{11}{21}.(n+2)(n+1)n(2n+1)(2n)​=2111​.

Cancel nnn: 2(2n+1)(n+2)(n+1)=1121.\frac{2(2n+1)}{(n+2)(n+1)}=\frac{11}{21}.(n+2)(n+1)2(2n+1)​=2111​.

  1. Cross-multiply: 42(2n+1)=11(n+2)(n+1).42(2n+1)=11(n+2)(n+1).42(2n+1)=11(n+2)(n+1).

Expand: 84n+42=11(n2+3n+2).84n+42=11(n^2+3n+2).84n+42=11(n2+3n+2). 84n+42=11n2+33n+22.84n+42=11n^2+33n+22.84n+42=11n2+33n+22.

Bring all terms to one side: 11n2−51n−20=0.11n^2-51n-20=0.11n2−51n−20=0.

  1. Solve the quadratic: 11n2−55n+4n−20=011n^2-55n+4n-20=011n2−55n+4n−20=0 11n(n−5)+4(n−5)=011n(n-5)+4(n-5)=011n(n−5)+4(n−5)=0 (11n+4)(n−5)=0.(11n+4)(n-5)=0.(11n+4)(n−5)=0.

So, n=5n=5n=5 (since nnn must be a positive integer for permutations).

  1. Now compute: n2+n+15=52+5+15=25+5+15=45.n^2+n+15=5^2+5+15=25+5+15=45.n2+n+15=52+5+15=25+5+15=45.

Therefore, the required value is 45.\boxed{45}.45​.

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