If $^{22} P_{r + 2} :^{22} P_{r + 1} = 11 : 55,$ find $r$ .

Open in App

Solution

$^{22} P_{r + 2} :^{22} P_{r + 1} = 11 : 55$

$\frac{22!}{(r + 2) (r + 1)!} \times \frac{(r + 1)!}{22!} = \frac{11}{55},$

$11 r + 22 = 55,$
$11 r = 33,$
$r = 3.$

Suggest Corrections

Similar questions

{^{56} P}_{r + 6} : {^{54} P}_{r + 3} = 30800 : 1

r

^{56} P_{R + 6} :^{54} P_{R + 3} = 30800 : 1

^{R} P_{2} .

^{5} P_{r} = 2^{6} P_{r - 1}

r

{^{5} P}_{r} = 2 {^{6} P}_{r - 1}

{^{5} P}_{r} = {^{6} P}_{r - 1}

\frac{p^{r}}{⌊ r} + \frac{p^{r - 1}}{⌊ r - 1} \cdot \frac{q}{1} + \frac{p^{r - 2}}{⌊ r - 2} \cdot \frac{q^{2}}{⌊ 2} + \frac{p^{r - 3}}{⌊ r - 3} \cdot \frac{q^{3}}{⌊ 3} + . . . . .

r + 1

Join BYJU'S Learning Program