Question

if
20 Pr
= 6840 findr.​

Answer 1

Correct Statement is

$\bf \: If \: \:^{20} P_r \: = \: 6840 \: then \: find \: value \: of \: r.$

$\large\underline\purple \bf{ \bf \: Solution :- }$

We know that,

• A permutation is the choice of r things from a set of n things without replacement and where the order matters.

$\rm :\implies\:\boxed{\tt\:^{n}P_r=\dfrac{n!}{(n-r)!}}$

So,

• Given that

$\rm :\longmapsto\:\:^{20} P_r \: = \: 6840$

$\rm :\longmapsto\:\dfrac{20!}{(20 - r)!} = 6840$

$\rm :\longmapsto\:\dfrac{ \cancel{20 \times 19 \times 18} \times 17!}{(20 - r)!} = \cancel{20 \times 19 \times 18}$

$\rm :\longmapsto\:\dfrac{17!}{(20 - r)!} = 1$

$\rm :\longmapsto\:17! = (20 - r)!$

$\rm :\implies\:17 = 20 - r$

$\rm :\implies\: \boxed{ \bf \: r \: = \: 3}$

Additional Information :-

Permutation :-

• A permutation is the choice of r things from a set of n things without replacement and where the order matters.

$\rm :\longmapsto\:\boxed{\bf\:^{n}P_r=\dfrac{n!}{(n-r)!}}$

Combination :-

• A combination is the choice of r things from a set of n things without replacement and where order does not matter.

$\rm :\longmapsto\:\boxed{\bf \:^n C_r=\dfrac{n!}{r! \: (n-r)!}}$