Question

if.npr=1320,ncr=220,find n and r

Answer 1

Heya user,

nPr = 1320 

and nCr = nPr / r! = 220
=> r! = nPr / 220  = 1320 / 220 = 6
=> r = 3;

Now, n! / ( n - 3 )! = 1320 
=> n ( n - 1 )( n - 2 ) = 1320

Let ( n - 1 ) = k;
Then, k(k+1)(k-1) = 1320
=> k³ - k = 1320
Since k is an integer, we see k = 11

And hence, (n-1) = 11 or n= 12

.'. n = 12; r = 3;

Answer 2

If npr=1320,ncr=220, the values of n and r are 12 and 3 respectively.

Given:

$P_{r} ^{n} = 1320$

$C_{r} ^{n} = 230$

To find:

Values of n and r = ?

Formula to be used:

$P_{r} ^{n} = \frac{n!}{(n-r)!}$

$C_{r} ^{n} = \frac{n!}{r!(n-r)!}$

Solution:

$P_{r} ^{n} = \frac{n!}{(n-r)!} = 1320\\\\C_{r} ^{n} = \frac{n!}{(n-r)!r!} = 220$

Dividing both equation:

$r! = \frac{11}{2}\\ \\r! = 5.5$

Rounding 5.5 to 6 we get,

r! = 6

As we know. 3! = 6

Therefore, r = 3

Substituting value or r in equation:

$\frac{n!}{(n-3)!} = 1320\\\\\frac{n(n-1)(n-2)(n-3)!}{(n-3)!} = 1320\\\\n(n-1)(n-2) = 1320$

Assume value of n to be 12, n-1=11 and n-2=10.

12(12-1)(12-2) = 12(11)(10) = 1320

Therefore, the values of r and n are 3 and 12 respectively.