Question

If $^6{P_r} = 360\,and{\,^6}{C_r} = 15$, then find r?

Answer 1

Hint: Apply the formula of permutation and combination, divide both the equations of permutation and combination, and find out the value of r.

Complete step-by-step answer:
Permutation: Arranging the numbers in order is called permutation, the formula of permutation is \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Where n= Total number of items in the sample, r= number of items to be selected from the sample.
Combination: Selecting the items from the sample is called combination, the formula of combination is \[^n{C_r} = {\dfrac{{n!}}{{r!\left( {n - r} \right)!}}_r}\]
Where n= Total number of items in the sample, r= number of items to be selected from the sample.
Now, given that $^6{P_r} = 360$So,
$
  { \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} \\
  { \Rightarrow ^6}{P_r} = \dfrac{{6!}}{{\left( {6 - r} \right)!}} = 360.......\left( 1 \right) \\
 $
And $^6{C_r} = 15$ can be written as
$
  { \Rightarrow ^n}{C_r} = {\dfrac{{n!}}{{r!\left( {n - r} \right)!}}_r} \\
  { \Rightarrow ^6}{C_r} = {\dfrac{{6!}}{{r!\left( {6 - r} \right)!}}_r} = 15........\left( 2 \right) \\
 $
Now, divide equation (1) with equation (2), we have
\[
   \Rightarrow \dfrac{{\left( {\dfrac{{6!}}{{\left( {6 - r} \right)!}}} \right)}}{{\left( {\dfrac{{6!}}{{r!\left( {6 - r} \right)!}}} \right)}} = \dfrac{{360}}{{15}} \\
   \Rightarrow \dfrac{{6!}}{{\left( {6 - r} \right)!}} \times \dfrac{{r!(6 - r)!}}{{6!}} = 24 \\
   \Rightarrow r! = 24 \\
 \]
Factorial of a number whose value 24 is 4. $4! = 4 \times 3 \times 2 \times 1 = 24$.
So, the value of r is 4.

Note: Don’t try to complicate the solution with elaborating the equation, divide the equations to simplify the calculations. A Permutation is an ordered Combination.