Question

(iv) If p(10,r) ==5040 then find the value of r​

Answer 1

$\large\underline{\sf{Given- }}$

$\sf \: \: P(10,r) \: = \: 5040$

$\large\underline{\sf{To\:Find - }}$

$\sf \: \: the \: value \: of \: r$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{\sf \: P(n, \: r) \: = \: \:^{n}P_r=\dfrac{n!}{(n-r)!}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: P(10,r) \: = \: 5040$

$\rm :\longmapsto\:\dfrac{10!}{(10 - r)!} = 5040$

$\rm :\longmapsto\:\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(10 - r)!} = 5040$

$\rm :\longmapsto\:\dfrac{\cancel{5040 }\times 6!}{(10 - r)!} = \cancel{5040}$

$\rm :\longmapsto\:(10 - r)! = 6!$

So, on comparing,

$\rm :\longmapsto\:10 - r = 6$

$\bf\implies \:r \: = \: 4$

Additional Information :-

$1. \: \: \: \boxed{\sf \:^{n}C_r=\dfrac{n!}{(n-r)!\times r!}}$

$2. \: \: \: \boxed{\sf \:^{n}C_r=\dfrac{1}{ r!} \times P(n, \: r)}$

$3. \: \: \: \boxed{\sf \:^{n}P_r=\dfrac{n}{r} \: \: ^{n - 1}P_{r - 1}}$

$4. \: \: \: \boxed{ \sf \: ^{n}C_r \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_r}$