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Question: If \[{}^{n}P_{r}=5040\] and \[{}^{n}C_{r}=210\], then find n and r....

If nPr=5040{}^{n}P_{r}=5040 and nCr=210{}^{n}C_{r}=210, then find n and r.

Explanation

Solution

Hint: In this question it is given that we have to find the value of n and r, where nPr=5040{}^{n}P_{r}=5040 and nCr=210{}^{n}C_{r}=210. So fo find the solution we need to know that nCr{}^{n}C_{r} can be written as, nCr=n!r!⋅(n−r)!{}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}
And also nPr{}^{n}P_{r} can be written as nPr=n!(n−r)!{}^{n}P_{r}=\dfrac{n!}{\left( n-r\right) !}.
Where, n!=n⋅(n−1)⋅(n−2)⋯3⋅2⋅1n!=n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdots 3\cdot 2\cdot 1

Complete step-by-step solution:
Given that,
nPr=5040{}^{n}P_{r}=5040.............(1)
Now as we know that nPr=n!(n−r)!{}^{n}P_{r}=\dfrac{n!}{\left( n-r\right) !}
So from equation (1), we can write,
n!(n−r)!=5040\dfrac{n!}{\left( n-r\right) !} =5040............(2)
Also here it is given,
nCr=210{}^{n}C_{r}=210
⇒n!r!⋅(n−r)!=210\Rightarrow \dfrac{n!}{r!\cdot \left( n-r\right) !} =210 [nCr=n!r!⋅(n−r)!{}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !} ]
⇒1r!×n!(n−r)!=210\Rightarrow \dfrac{1}{r!} \times \dfrac{n!}{\left( n-r\right) !} =210
Now putting the value of n!(n−r)!\dfrac{n!}{\left( n-r\right) !} from the equation (2), we can write the above equation as,
1r!×5040=210\dfrac{1}{r!} \times 5040=210
⇒1r!=2105040\Rightarrow \dfrac{1}{r!} =\dfrac{210}{5040} [dividing both side by 5040]
⇒1r!=124\Rightarrow \dfrac{1}{r!} =\dfrac{1}{24}
Now by cross multiplication the above equation can be written as,
⇒r!=24\Rightarrow r!=24
⇒r!=4×3×2×1\Rightarrow r!=4\times 3\times 2\times 1
⇒r!=4!\Rightarrow r!=4!
Therefore r=4,
Now by putting the value of r in equation (2) we get,
n!(n−4)!=5040\dfrac{n!}{\left( n-4\right) !} =5040
Now since n!=n
⇒n⋅(n−1)⋅(n−2)⋅(n−3)⋅(n−4)!(n−4)!=5040\Rightarrow \dfrac{n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdot \left( n-3\right) \cdot \left( n-4\right) !}{\left( n-4\right) !} =5040
⇒n(n−1)(n−2)(n−3)=5040\Rightarrow n\left( n-1\right) \left( n-2\right) \left( n-3\right) =5040...........(3)
Now we are going to factorise 5040 and we have to express it the multiplication of four terms,
5040=7×6×5×4×3×2×15040=7\times 6\times 5\times 4\times 3\times 2\times 1
=7×6×5×4×3×2=7\times 6\times 5\times 4\times 3\times 2
=7×(3×2)×5×4×3×2=7\times \left( 3\times 2\right) \times 5\times 4\times 3\times 2
=7×3×2×5×4×3×2=7\times 3\times 2\times 5\times 4\times 3\times 2
=7×(2×4)×(5×2)×(3×3)=7\times (2\times 4)\times (5\times 2)\times (3\times 3) [by rearranging]
=7×8×10×9=7\times 8\times 10\times 9
=10×9×8×7=10\times 9\times 8\times 7
Now putting the factored form of 5040 in the equation (3) we get,
n⋅(n−1)⋅(n−2)⋅(n−3)=10×9×8×7n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdot \left( n-3\right) =10\times 9\times 8\times 7
n⋅(n−1)⋅(n−2)⋅(n−3)=10⋅(10−1)⋅(10−2)⋅(10−3)n\cdot \left( n-1\right) \cdot \left( n-2\right) \cdot \left( n-3\right) =10\cdot \left( 10-1\right) \cdot \left( 10-2\right) \cdot \left( 10-3\right)
Now comparing the both sides of the above equation, we can easily say that the value of n is 10, i.e, n = 10.
Therefore r= 4 and n= 10.
Which is our required solution.

Note: While solving this type of question you need to know that we cannot solve the above equation by conventional method, so that is why we use the comparison method, i.e we have made the same structure on both sides of the equation and after that we can equate the corresponding values.