Question

If p(n ,r)=2520 and c(n, r)=21 then what is the value of c (n+1, r+1)?
a)7
b)14
c)28
d)56

Answer 1

Hint: This is related to permutation and combinations. So directly go for its formulas.
Formulas used:
1.\[p(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
2.\[c(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step-by-step answer:
We are given p (n ,r)=2520 but using formula 1 we can write
\[\dfrac{{n!}}{{\left( {n - r} \right)!}}\]=2520 ………equation 1
Also we are given c (n, r)=21 and using formula 2 we can write
\[\dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]=21 …………….equation 2
Replacing \[\dfrac{{n!}}{{\left( {n - r} \right)!}}\] of equation 2 with 2520 from equation 1 we get,
\[
  \dfrac{{2520}}{{r!}} = 21 \\
\Rightarrow r! = \dfrac{{2520}}{{21}} \\
\Rightarrow r! = 120 \\
\Rightarrow r = 5 \\
\]
Thus r=5.
Now ,
\[p(n,5) = \dfrac{{n!}}{{\left( {n - 5} \right)!}}\]
\[2520 = \dfrac{{n!}}{{\left( {n - 5} \right)!}}\]
\[2520 = n(n - 1)(n - 2)(n - 3)(n - 4)\]
\[n = 7\]
Thus n=7.
Now,
\[c(8,6) = \dfrac{{8!}}{{6!\left( {8 - 6} \right)!}} = \dfrac{{8!}}{{6! \times 2!}} = \dfrac{{56}}{2} = 28\]
Thus, option c is the correct answer.

Note: There are two different formulas for permutations and combinations.
1.Here n=7 is the answer because 6!=120 but this is less than 2520. And 7!=5040 but it is greater than 2520.
2.If you observe the last term in the numerator is (n-4). So the number is divided by 2 [n-5=7-5=2 is not used].
3.In the last step of c(8, 6) don’t calculate all factorials. First cancel common factors or lesser factorials.
For example 8! And 6! have 6! common .Then just calculate the remaining product.