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Question: The expression \({}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5...

The expression 45C8+∑k=1752−kC7+∑i=1557−iC50−i{}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}}
A. 55C7{}^{55}{C_7}
B. 57C8{}^{57}{C_8}
C. 57C7{}^{57}{C_7}
D. None of these

Explanation

Solution

We can expand the 2nd term of the expression by giving values for k. We can simplify the last term using the relation nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}. Then we get a series of combinations whose 1st two consecutive terms can be simplified using the relation nCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}. Then the resulting series can also be simplified as above and this continues until we get the expression in a single term.

Complete step by step Answer:

We have the expression 45C8+∑k=1752−kC7+∑i=1557−iC50−i{}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}} . Let the value of the expression be II
⇒I=45C8+∑k=1752−kC7+∑i=1557−iC50−i\Rightarrow I = {}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}}
We know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}} , we can use this in the 3rd term of the expression.
⇒∑i=1557−iC50−i=∑i=1557−iC57−i−50+i\Rightarrow\sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}} = \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{57 - i - 50 + i}}}
On simplification, we get,
⇒∑i=1557−iC50−i=∑i=1557−iC7\Rightarrow\sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}} = \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_7}}
So, the expression becomes,
45C8+∑k=1752−kC7+∑i=1557−iC7{}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_7}} …………..(A)
Now we can expand the summations to get a series,
On expanding the 2nd term of the expression, we get,
⇒∑k=1752−kC7=52−1C7+52−2C7+....+52−6C7+52−7C7\Rightarrow\sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} = {}^{52 - 1}{C_7} + {}^{52 - 2}{C_7} + .... + {}^{52 - 6}{C_7} + {}^{52 - 7}{C_7}
On simplifying and rearranging, we get,
⇒∑k=1752−kC7=45C7+46C7+....+50C7+51C7\Rightarrow\sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} = {}^{45}{C_7} + {}^{46}{C_7} + .... + {}^{50}{C_7} + {}^{51}{C_7} … (1)
On expanding the 3rd term of expression(A), we get,
⇒∑i=1557−iC7=57−1C7+57−2C7+...+57−4C7+57−5C7\Rightarrow \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_7}} = {}^{57 - 1}{C_7} + {}^{57 - 2}{C_7} + ... + {}^{57 - 4}{C_7} + {}^{57 - 5}{C_7}
On simplifying and rearranging, we get,
⇒∑i=1557−iC7=52C7+53C7+...+55C7+56C7\Rightarrow \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_7}} = {}^{52}{C_7} + {}^{53}{C_7} + ... + {}^{55}{C_7} + {}^{56}{C_7}… (2)
We can substitute (1) and (2) in the expression.
⇒I=45C8+45C7+46C7+....+50C7+51C7+52C7+53C7+...+55C7+56C7\Rightarrow I = {}^{45}{C_8} + {}^{45}{C_7} + {}^{46}{C_7} + .... + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + ... + {}^{55}{C_7} + {}^{56}{C_7}
We know that nCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}. We can apply this relation to the 1st two terms of the series.
⇒45C8+45C7=46C8\Rightarrow{}^{45}{C_8} + {}^{45}{C_7} = {}^{46}{C_8}
Now the expression becomes,
⇒I=46C8+46C7+....+50C7+51C7+52C7+53C7+...+55C7+56C7\Rightarrow I = {}^{46}{C_8} + {}^{46}{C_7} + .... + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + ... + {}^{55}{C_7} + {}^{56}{C_7}
The same relation can be applied in the expression and this process continues.
Thus, in the end, we will get,
⇒I=56C8+56C7=57C8\Rightarrow I = {}^{56}{C_8} + {}^{56}{C_7} = {}^{57}{C_8}
Therefore, the value of the expression is 57C8{}^{57}{C_8}
So, the correct answer is option B.

Note: The concept of combinations and properties are used to solve this problem. The symbol ∑\sum {} represents a summation operator. It is the process of taking the sum or adding up all the terms coming upper the operation. Here we expanded all the terms under this operation and arranged them in an order which is easy for calculations. We don’t need to expand the combinations to simplify the series. We can use the relation nCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}to simplify the series. As there are many terms in the series, we cannot solve it one by one. So, we can simplify the few terms and the trend will continue till the last term.