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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

First we can see that the last terms has common term I in place of both n and r in nCr{}^n{C_r} so we can use the formula nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}} to simplify the term. Then open the summation and put the values of k and i from 11 to 77 and 11 to 55 respectively. Then rearrange the terms in increasing order of n. Then we can use the rulenCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r} again and again till we get the last term.

Complete step-by-step answer:
Given expression is45C8+∑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}}} --- (i)
We know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}
So we can write-
⇒57−iC50−i=57−iC57−i−(50−i)\Rightarrow {}^{57 - i}{C_{50 - i}} = {}^{57 - i}{C_{57 - i - \left( {50 - i} \right)}}
On solving, we get-
⇒57−iC50−i=57−iC57−i−50+i\Rightarrow {}^{57 - i}{C_{50 - i}} = {}^{57 - i}{C_{57 - i - 50 + i}}
On further solving, we get-
⇒57−iC50−i=57−iC57−50\Rightarrow {}^{57 - i}{C_{50 - i}} = {}^{57 - i}{C_{57 - 50}}
On subtraction, we get-
⇒57−iC50−i=57−iC7\Rightarrow {}^{57 - i}{C_{50 - i}} = {}^{57 - i}{C_7}
On putting this value in eq. (i), we get-
⇒45C8+∑k=1752−kC7+∑i=1557−iC7\Rightarrow {}^{45}{C_8} + \sum\nolimits_{k = 1}^7 {{}^{52 - k}{C_7}} + \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_7}}
On putting the value of k and i in the equation and removing the summation sign, we get-
⇒45C8+52−1C7+52−2C7+52−3C7+52−4C7+52−5C7+52−6C7+52−7C7+57−1C7+57−2C7+57−3C7+57−4C7+57−5C7\Rightarrow {}^{45}{C_8} + {}^{52 - 1}{C_7} + {}^{52 - 2}{C_7} + {}^{52 - 3}{C_7} + {}^{52 - 4}{C_7} + {}^{52 - 5}{C_7} + {}^{52 - 6}{C_7} + {}^{52 - 7}{C_7} + {}^{57 - 1}{C_7} + {}^{57 - 2}{C_7} + {}^{57 - 3}{C_7} + {}^{57 - 4}{C_7} + {}^{57 - 5}{C_7}On simplifying, we get-
⇒45C8+51C7+50C7+49C7+48C7+47C7+46C7+45C7+56C7+55C7+54C7+53C7+52C7\Rightarrow {}^{45}{C_8} + {}^{51}{C_7} + {}^{50}{C_7} + {}^{49}{C_7} + {}^{48}{C_7} + {}^{47}{C_7} + {}^{46}{C_7} + {}^{45}{C_7} + {}^{56}{C_7} + {}^{55}{C_7} + {}^{54}{C_7} + {}^{53}{C_7} + {}^{52}{C_7}
On re-arranging, we can write-
⇒45C8+45C7+51C7+50C7+49C7+48C7+47C7+46C7+56C7+55C7+54C7+53C7+52C7\Rightarrow {}^{45}{C_8} + {}^{45}{C_7} + {}^{51}{C_7} + {}^{50}{C_7} + {}^{49}{C_7} + {}^{48}{C_7} + {}^{47}{C_7} + {}^{46}{C_7} + {}^{56}{C_7} + {}^{55}{C_7} + {}^{54}{C_7} + {}^{53}{C_7} + {}^{52}{C_7}
Now, we know that nCr+nCr−1=n+1Cr{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r} -- (ii)
On applying this on the first two terms, we get-
⇒45+1C8+51C7+50C7+49C7+48C7+47C7+46C7+56C7+55C7+54C7+53C7+52C7\Rightarrow {}^{45 + 1}{C_8} + {}^{51}{C_7} + {}^{50}{C_7} + {}^{49}{C_7} + {}^{48}{C_7} + {}^{47}{C_7} + {}^{46}{C_7} + {}^{56}{C_7} + {}^{55}{C_7} + {}^{54}{C_7} + {}^{53}{C_7} + {}^{52}{C_7}
On solving, we get-
⇒46C8+51C7+50C7+49C7+48C7+47C7+46C7+56C7+55C7+54C7+53C7+52C7\Rightarrow {}^{46}{C_8} + {}^{51}{C_7} + {}^{50}{C_7} + {}^{49}{C_7} + {}^{48}{C_7} + {}^{47}{C_7} + {}^{46}{C_7} + {}^{56}{C_7} + {}^{55}{C_7} + {}^{54}{C_7} + {}^{53}{C_7} + {}^{52}{C_7}
Again rearranging, we get-
⇒46C8+46C7+47C7+48C7+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{46}{C_8} + {}^{46}{C_7} + {}^{47}{C_7} + {}^{48}{C_7} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
Now again applying the rule of eq. (ii) in the first two terms, we get-
⇒46+1C8+47C7+48C7+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{46 + 1}{C_8} + {}^{47}{C_7} + {}^{48}{C_7} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On solving, we get-
⇒47C8+47C7+48C7+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{47}{C_8} + {}^{47}{C_7} + {}^{48}{C_7} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
Again applying the rule (ii) in the above eq. we get-
⇒47+1C8+48C7+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{47 + 1}{C_8} + {}^{48}{C_7} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On addition, we get-
⇒48C8+48C7+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{48}{C_8} + {}^{48}{C_7} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
Again applying the same rule, we get-
⇒48+1C8+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{48 + 1}{C_8} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On addition, we get-
⇒49C8+49C7+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{49}{C_8} + {}^{49}{C_7} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On again applying the same rule, we get-
⇒49+1C8+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{49 + 1}{C_8} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On solving, we get-
⇒50C8+50C7+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{50}{C_8} + {}^{50}{C_7} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
On again applying the rule of eq. (ii) in the above equation, we get-
⇒51C8+51C7+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{51}{C_8} + {}^{51}{C_7} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
Again reapplying the rule and adding, we get-
⇒52C8+52C7+53C7+54C7+55C7+56C7\Rightarrow {}^{52}{C_8} + {}^{52}{C_7} + {}^{53}{C_7} + {}^{54}{C_7} + {}^{55}{C_7} + {}^{56}{C_7}
Applying the same rule till we get only two terms, we get-
⇒56C8+56C7\Rightarrow {}^{56}{C_8} + {}^{56}{C_7}
Again applying the same rule, we get-
⇒57C8\Rightarrow {}^{57}{C_8}
Hence the correct answer is option B.

Note: Here, if we put the value of i directly without using the formula, we will get-
⇒∑i=1557−iC50−i=56C49+55C48+54C47+53C46+52C45\Rightarrow \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}} = {}^{56}{C_{49}} + {}^{55}{C_{48}} + {}^{54}{C_{47}} + {}^{53}{C_{46}} + {}^{52}{C_{45}}
Now since the values are too big we will have to use nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}} and then we will get-
⇒∑i=1557−iC50−i=56C7+55C7+54C7+53C7+52C7\Rightarrow \sum\nolimits_{i = 1}^5 {{}^{57 - i}{C_{50 - i}}} = {}^{56}{C_7} + {}^{55}{C_7} + {}^{54}{C_7} + {}^{53}{C_7} + {}^{52}{C_7}
So we get the same terms as we got on directly using this formula on the last term of expression. So students can also solve the question this way then proceed further as in the above solution.