Question

If ${}^n{P_r} = 840$ and ${}^n{C_r} = 35$, find the value of r.

Answer 1

Hint: We have 2 equations. We can expand the LHS of both the equations using the expansion of permutations and combinations. Then we can simplify the factorials. Then we will obtain 2 equations with 2 variables. Then we can solve them using appropriate substitution to get the required value of r.

Complete step-by-step answer:
We are given that ${}^n{P_r} = 840$.
We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ So, we can expand the LHS of the equation as,
 $ \Rightarrow \dfrac{{n!}}{{\left( {n - r} \right)!}} = 840$ …. (1)
We are also given that ${}^n{C_r} = 35$.
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. So, we can expand its LHS as,
 $ \Rightarrow \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 35$
We can write the LHS as,
 $ \Rightarrow \dfrac{1}{{r!}} \times \dfrac{{n!}}{{\left( {n - r} \right)!}} = 35$
Now we can substitute equation (1).
 $ \Rightarrow \dfrac{1}{{r!}} \times 840 = 35$
On rearranging, we get,
 $ \Rightarrow r! = \dfrac{{840}}{{35}}$
On simplification, we get,
 $ \Rightarrow r! = 24$
Now we can factorise the RHS.
 $ \Rightarrow r! = 4 \times 3 \times 2 \times 1$
Now the RHS is equal to $4!$. So, we can write,
 $ \Rightarrow r! = 4!$
Hence, we have,
 $ \Rightarrow r = 4$
Therefore, the required value of r is 4.

Note: We must note that the expansion of combinations is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and expansion of permutations is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. We cannot give substitution for only one variable as the variables inside the factorials cannot be separated. We must also note that $\left( {a - b} \right)! \ne a! - b!$.
Alternate method to find the solution is given by,
We are given that ${}^n{P_r} = 840$ and we are also given that ${}^n{C_r} = 35$.
We know that the relation between permutations and combinations is given by the equation,
 ${}^n{C_r} = \dfrac{{{}^n{P_r}}}{{r!}}$
On rearranging, we get,
 $ \Rightarrow r! = \dfrac{{{}^n{P_r}}}{{{}^n{C_r}}}$
On substituting the given values, we get,
 $ \Rightarrow r! = \dfrac{{840}}{{35}}$
On simplification, we get,
 $ \Rightarrow r! = 24$
Now we can factorise the RHS.
 $ \Rightarrow r! = 4 \times 3 \times 2 \times 1$
Now the RHS is equal to $4!$. So, we can write,
 $ \Rightarrow r! = 4!$
Hence, we have,
 $ \Rightarrow r = 4$
Therefore, the required value of r is 4.