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Question: If \(^6{P_r} = 360\,and{\,^6}{C_r} = 15\), then find r?...

If 6Pr=360and6Cr=15^6{P_r} = 360\,and{\,^6}{C_r} = 15, then find r?

Explanation

Solution

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 nPr=n!(nr)!^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 nCr=n!r!(nr)!r^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 6Pr=360^6{P_r} = 360So,
nPr=n!(nr)! 6Pr=6!(6r)!=360.......(1)  { \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 6Cr=15^6{C_r} = 15 can be written as
nCr=n!r!(nr)!r 6Cr=6!r!(6r)!r=15........(2)  { \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

(6!(6r)!)(6!r!(6r)!)=36015 6!(6r)!×r!(6r)!6!=24 r!=24  \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!=4×3×2×1=244! = 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.