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Question: The value of \[{}^{40}{C_{31}} + \sum\limits_{j = 0}^{10} {{}^{40 + j}{C_{10 + j}}} \] is equal to ...

The value of 40C31+∑j=01040+jC10+j{}^{40}{C_{31}} + \sum\limits_{j = 0}^{10} {{}^{40 + j}{C_{10 + j}}} is equal to
A. 51C20{}^{51}{C_{20}}
B. 2⋅50C202 \cdot {}^{50}{C_{20}}
C. 2⋅45C152 \cdot {}^{45}{C_{15}}
D. None of these

Explanation

Solution

As we know combination determines the number of possible arrangements in a collection of items in which the order of the selection does not matter and hence, to solve the given equation of combination, apply the formula to get the combined terms and the formula is given as nCr−1+nCr=n+1Cr{}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}.

Formula used:
nCr−1+nCr=n+1Cr{}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}
Where,
nn= is number of items.
rr= number of items are taken at a time.
The number of combinations of ‘n’ different things taken ‘r’ at a time is denoted by nCr{}^n{C_r}, in which
nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

Complete step-by-step answer:
To find the value of the above sum given
40C31+∑j=01040+jC10+j{}^{40}{C_{31}} + \sum\limits_{j = 0}^{10} {{}^{40 + j}{C_{10 + j}}}
Let us expand the sum using the combination formula
nCr−1+nCr=n+1Cr{}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}
In which as per the sum given here, n=40n = 40
Expanding the terms with respect to values
40C31+[40C10+41C11+42C12+..........50C20]{}^{40}{C_{31}} + \left[ {{}^{40}{C_{10}} + {}^{41}{C_{11}} + {}^{42}{C_{12}} + ..........{}^{50}{C_{20}}} \right]
=40C9+[40C10+41C11+42C12+..........50C20]= {}^{40}{C_9} + \left[ {{}^{40}{C_{10}} + {}^{41}{C_{11}} + {}^{42}{C_{12}} + ..........{}^{50}{C_{20}}} \right]
=(40C9+40C10)+41C11+42C12+..........50C20= \left( {{}^{40}{C_9} + {}^{40}{C_{10}}} \right) + {}^{41}{C_{11}} + {}^{42}{C_{12}} + ..........{}^{50}{C_{20}}
=(41C10+41C11)+42C12+..........50C20= \left( {{}^{41}{C_{10}} + {}^{41}{C_{11}}} \right) + {}^{42}{C_{12}} + ..........{}^{50}{C_{20}}
=40C11+42C12+..........50C20= {}^{40}{C_{11}} + {}^{42}{C_{12}} + ..........{}^{50}{C_{20}}
Further simplifying the terms, we get
=49C18+49C19+50C20= {}^{49}{C_{18}} + {}^{49}{C_{19}} + {}^{50}{C_{20}}
=50C19+50C20= {}^{50}{C_{19}} + {}^{50}{C_{20}}
Hence, after simplification we get
=51C20= {}^{51}{C_{20}}
So, the correct answer is “Option A”.

Note: The key point to a combination is that there are no repetitions of objects allowed and order is not important to find a combination. In the same way we can find the number of Permutations of ′n′'n' different things taken ′r′'r'at a time is denoted by nPr{}^n{P_r} , in which
nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}.