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Question: If \(n{C_r} = 10\), \(n{C_{r + 1}} = 45\) then, \(r\) equals to A. 1 B. 2 C. 3 D. 4...

If nCr=10n{C_r} = 10, nCr+1=45n{C_{r + 1}} = 45 then, rr equals to
A. 1
B. 2
C. 3
D. 4

Explanation

Solution

In this problem, we have given the value of nn number of different items and we have to choose rr number of items from it. And also we have given the value of nn number of different items and we have to choose r+1r + 1 number of items from it. Here our aim is to find the value of rr. To solve this problem we are going to use the combination formula and by using this formula we find the value of rr.

Formula used: nCr=n!r!(n−r)!n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Where,
nn= the number of items and
rr=how many items are taken at a time.
nCr+1=n!(r+1)!(n−r−1)!n{C_{r + 1}} = \dfrac{{n!}}{{(r + 1)!(n - r - 1)!}}
Where,
nn= the number of items and
r+1r + 1=how many items are taken at a time.

Complete step-by-step solution:
Here it is given in the question that, nCr=10n{C_r} = 10, nCr+1=45n{C_{r + 1}} = 45
That is, nCr=n!r!(n−r)!=10n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 10
And nCr+1=n!(r+1)!(n−r−1)!=45n{C_{r + 1}} = \dfrac{{n!}}{{(r + 1)!(n - r - 1)!}} = 45
Let us consider that,
nCr+1nCr=n!(r+1)!(n−r−1)!n!r!(n−r)!\dfrac{{n{C_{r + 1}}}}{{n{C_r}}} = \dfrac{{\dfrac{{n!}}{{\left( {r + 1} \right)!(n - r - 1)!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}}
Hence we get,
⇒n!(r+1)!(n−r−1)!n!r!(n−r)!=4510\Rightarrow \dfrac{{\dfrac{{n!}}{{\left( {r + 1} \right)!(n - r - 1)!}}}}{{\dfrac{{n!}}{{r!(n - r)!}}}} = \dfrac{{45}}{{10}}
Solving this we get,
⇒n−rr+1=4510−−−−−(∗)\Rightarrow \dfrac{{n - r}}{{r + 1}} = \dfrac{{45}}{{10}} - - - - - \left( * \right)
Solving the numerical fraction in the right hand side, we get
⇒n−rr+1=92\Rightarrow \dfrac{{n - r}}{{r + 1}} = \dfrac{9}{2}
Cross multiply the above equation we get,
2(n−r)=9(r+1)2(n - r) = 9(r + 1)
2n−2r=9r+92n - 2r = 9r + 9
Keep the r terms in one side, we get
2n=11r+9−−−−−(1)2n = 11r + 9 - - - - - (1)
Now let’s give the values for rr
If r=2 or 4r = 2{\text{ or }}4 then equation 1 becomes,
r=2⇒n=312r = 2 \Rightarrow n = \dfrac{{31}}{2} is not an integer.
r=4⇒n=532r = 4 \Rightarrow n = \dfrac{{53}}{2} is not an integer.
If r=3r = 3 then n=422=21n = \dfrac{{42}}{2} = 21
So in the case r=3r = 3 we get n=21n = 21, and nCr=21C3≠10n{C_r} = 21{C_3} \ne 10
So r=3r = 3 is not a correct answer.
If r=1⇒n=202=10r = 1 \Rightarrow n = \dfrac{{20}}{2} = 10 which is an integer
Then in the case when r=1&n=10r = 1\& n = 10 the combination formula nCr=10C1=10n{C_r} = 10{C_1} = 10 and also nCr+1=10C2=45n{C_{r + 1}} = 10{C_2} = 45
Therefore, r=1r = 1 and n=10n = 10 is the correct answer.

Hence, the answer is option (A)

Note: We have to remember that, in mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter (unlike permutations).
We can write (r+1)!\left( {r + 1} \right)!cas r!(r+1)r!(r + 1)also we can write (n−r)!\left( {n - r} \right)! as (n−r)(n−r−1)!(n - r)(n - r - 1)!. This is the main thing we used in our problem to bring up the required solution. Equation (*) is claimed by using these formulas and cancelling all the same terms in the numerator and in the denominator. And also we substituted the value of r and n in the nCrn{C_r} and nCr+1n{C_{r + 1}} formulas to get the given required solution.