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Question: Find \(n\) and \(r\), if \({}^{n}{{P}_{r}}=720\) and \({}^{n}{{C}_{r}}=120\)....

Find nn and rr, if nPr=720{}^{n}{{P}_{r}}=720 and nCr=120{}^{n}{{C}_{r}}=120.

Explanation

Solution

The given question is based on permutations and combinations. In order to solve this question we need to apply the formulas of permutation and combinations, which are given by
nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Where, n=n= total number of objects and
r=r= number of objects to be selected.
Then, substituting the values given in the question and solving further we get the desired answer.

Complete step-by-step answer:
We have been given that nPr=720{}^{n}{{P}_{r}}=720 and nCr=120{}^{n}{{C}_{r}}=120.
We have to find the values of nn and rr.
Now, we know that the formula of permutation is given by nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
Substituting the value, we have 720=n!(n−r)!..........(i)720=\dfrac{n!}{\left( n-r \right)!}..........(i)
Also, we know that the formula of combination is given by nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Substituting the value, we have 120=n!r!(n−r)!120=\dfrac{n!}{r!\left( n-r \right)!}
Or we can write that 120×r!=n!(n−r)!............(ii)120\times r!=\dfrac{n!}{\left( n-r \right)!}............(ii)
Now, when we compare equation (i) and equation (ii), we analyze that R.H.S. of both the equations is equal so the L.H.S. of both the equations will also be equal,
So, we have 120×r!=720120\times r!=720
Or we can write that
r!=720120 r!=6 \begin{aligned} & r!=\dfrac{720}{120} \\\ & r!=6 \\\ \end{aligned}
Now, we can write 66 as 3×2×13\times 2\times 1 which is equal to 3!3!,
So, we have r!=3!r!=3!
Or r=3r=3 .
Now, substituting the value of rr into equation (i), we get
720=n!(n−3)!720=\dfrac{n!}{\left( n-3 \right)!}
Now, we know that we can write n!=n(n−1)(n−2)(n−3)!n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!
Now, we have 720=n(n−1)(n−2)(n−3)!(n−3)!720=\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)!}{\left( n-3 \right)!}
On solving we get
720=n(n−1)(n−2)720=n\left( n-1 \right)\left( n-2 \right)
Now, we can factorize 720720 as 10×9×810\times 9\times 8
So, we have n(n−1)(n−2)=10×9×8n\left( n-1 \right)\left( n-2 \right)=10\times 9\times 8
On comparing we get n=10n=10.
The value of r=3r=3 and n=10n=10.

Note: Alternatively one can directly calculate the value of rr by using the relation nPr=nCr×r!{}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!.
Substituting the values and solving further to get the value of rr and solving the remaining part as we solved above to get the value of nn.