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Question: If \(a\) denotes the permutations of \(\left( {x + 2} \right)\) things taken all at a time, \(b\) th...

If aa denotes the permutations of (x+2)\left( {x + 2} \right) things taken all at a time, bb the number of permutations of xx things taken 11 at a time and cc the number of permutations of x−11x - 11 things taken all at a time such that a=182a = 182 bcbc, find the value of xx.
AA. (12,15)\left( {12,15} \right)
BB. (12,−20)\left( {12, - 20} \right)
CC. (22,−15)\left( {22, - 15} \right)
DD. (12,−35)\left( {12, - 35} \right)

Explanation

Solution

Hint: Here we will proceed by using the formula for permutations that is nPr=n!n−r!{}^n{P_r} = \dfrac{{n!}}{{n - r!}}. Then by applying the conditions given in the question we will get our answer.

Complete step-by-step solution -
If a denotes the number of permutations of (x+2)\left( {x + 2} \right) then, a=x+2Px+2a = {}^{x + 2}{P_{x + 2}}

B denotes the things taken 11 at a time,
b=xp11b = {}^x{p_{11}}

C denotes the number of permutations of x−11x - 11 taken all at a time
C=x−11Px−11C = {}^{x - 11}{P_{x - 11}}

It is given that, a=182a = 182 bcbc
x+2Px+2=182×xP11×x−11Px−11{}^{x + 2}{P_{x + 2}} = 182 \times {}^x{P_{11}} \times {}^{x - 11}{P_{x - 11}}
⇒(x+2)!0!=182×x!(x−11)!×(x−11)!0!\Rightarrow \dfrac{{\left( {x + 2} \right)!}}{{0!}} = 182 \times \dfrac{{x!}}{{\left( {x - 11} \right)!}} \times \dfrac{{\left( {x - 11} \right)!}}{{0!}}
We know that, 0! = 1, then

⇒x+2!=182×x! \Rightarrow x + 2! = 182 \times x!
⇒(x+2)(x+1)×x!=182×x!\Rightarrow \left( {x + 2} \right)\left( {x + 1} \right) \times x! = 182 \times x!

By using quadratic equation, we will get
⇒x2+3x+2=182\Rightarrow {x^2} + 3x + 2 = 182
⇒x2+3x−180=0\Rightarrow {x^2} + 3x - 180 = 0
Now by factorising we will get,
⇒x2+15x−12x−180=0 ⇒(x+15)(x−12)=0  \Rightarrow {x^2} + 15x - 12x - 180 = 0 \\\ \Rightarrow \left( {x + 15} \right)\left( {x - 12} \right) = 0 \\\
Therefore, x=−15,x=12x = - 15,x = 12

nPr{}^n{P_r} equation is always positive and even if we put the value of x as -15, the whole equation will become negative. Therefore, we will reject the negative value.
Hence, the correct answer is 12.

Note: Whenever we come up with this type of problem, one must think that permutations are an act of arranging the objects or numbers. Then by simplifying the equation by either using factoring by quadratic equations or any other method we can easily solve these types of questions.