Question

Find the value of n; if given that: \[^{n + 2}{C_8}{:^{n - 2}}{P_4} = 57:16\]
(a) $19$
(b) \[18\]
(c) Cannot be determined
(d) None of these

Answer 1

Hint: The given problem revolves around the concepts of permutations and combinations terminology. We use standard formulae for permutations and combinations of and form an equation and then make n as subject and solve to obtain its value.
Formula used
I.\[^n{P_r}\] equals to $\dfrac{{n!}}{{\left( {n - r} \right)!}}$
II. \[^n{C_r}\] equals to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where n is the total number of objects and r is the number of selected objects.

Complete step-by-step answer:
The given expression for the combinations of the equations can be written as, \[^{n + 2}{C_8}{:^{n - 2}}{P_4} = 57:16\]
Since, we know that
In case of permutations and combinations of any factorials that is in the form of \[^n{P_r}\] equals to $\dfrac{{n!}}{{\left( {n - r} \right)!}}$ and \[^n{C_r}\] equals to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ respectively,
Hence, considering \[^{n + 2}{C_8}\],
As a result, comparing and substituting the respective values, we get
\[^{n + 2}{C_8} = \dfrac{{\left( {n + 2} \right)!}}{{8!\left( {n + 2 - 8} \right)!}}\]
Solving the equation mathematically, we get
\[^{n + 2}{C_8} = \dfrac{{\left( {n + 2} \right)!}}{{8!\left( {n - 6} \right)!}}\] … (i)
Similarly,
Considering \[^{n - 2}{P_4}\],
As a result, comparing and substituting the respective values, we get
\[^{n - 2}{P_4} = \dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 2 - 4} \right)!}}\]
Solving the equation mathematically, we get
\[^{n - 2}{P_4} = \dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 6} \right)!}}\] … (ii)
From (i) and (ii),
Dividing both the equations, we get
$\dfrac{{^{n + 2}{C_8}}}{{^{n - 2}{P_4}}} = \dfrac{{\left( {\dfrac{{\left( {n + 2} \right)!}}{{8!\left( {n - 6} \right)!}}} \right)}}{{\left( {\dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 6} \right)!}}} \right)}}$
Solving the equation mathematically, we get
$\dfrac{{^{n + 2}{C_8}}}{{^{n - 2}{P_4}}} = \dfrac{{\left( {n + 2} \right)!}}{{8!}} \times \dfrac{1}{{\left( {n - 2} \right)!}}$
By using the definition of factorials, we get
 $\dfrac{{^{n + 2}{C_8}}}{{^{n - 2}{P_4}}} = \dfrac{{\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)!}}{{8!}} \times \dfrac{1}{{\left( {n - 2} \right)!}}$
$\dfrac{{^{n + 2}{C_8}}}{{^{n - 2}{P_4}}} = \dfrac{{\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)}}{{8!}}$
Since, we have given that
$\dfrac{{^{n + 2}{C_8}}}{{^{n - 2}{P_4}}} = \dfrac{{57}}{{16}}$
Hence, the equation becomes
\[\dfrac{{\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)}}{{8!}} = \dfrac{{57}}{{16}}\]
Simplifying the equation, we get,
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = \dfrac{{57}}{{16}} \times 8!\]
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = \dfrac{{57}}{{16}} \times 8!\]
Solving the equation predominantly, we get
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = \dfrac{{19 \times 3}}{{16}} \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\]
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = 19 \times 3 \times 7 \times 6 \times 5 \times 4 \times 3\]
Grouping the R.H.S. such that to equate all the four terms exists in L.H.S., we get
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = \left( {3 \times 7} \right)\left( {5 \times 4} \right)\left( {19} \right)\left( {6 \times 3} \right)\]
As a result, solving the equation, we get
\[\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right) = \left( {21} \right)\left( {20} \right)\left( {19} \right)\left( {18} \right)\]
Hence, simultaneously equating the L.H.S. and R.H.S., we get
\[n + 2 = 21 \to n = 21 - 2 = 19\] … (iii)
\[n + 1 = 20 \to n = 20 - 1 = 19\] … (iv)
\[n = 19\] … (v)
\[n - 1 = 18 \to n = 18 + 1 = 19\] … (vi)
Hence, from the above equations i.e. (iii), (iv), (v) and (vi)
It seems that, the required value of ${n^{th}}$ term is
$ \Rightarrow n = 19$.
So, the correct answer is “Option a”.

Note: Permutations can be defined as the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor. When order is not a factor it is called combination. Combination is the counting of selections that we make from n objects. Whereas permutation is counting the number of arrangements from n objects.