
Choose the correct value of \[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} \] from the following options:
A. \[{}^{50}{C_3}\]
B. \[{}^{56}{C_4}\]
C. \[{}^{55}{C_4}\]
D. \[{}^{55}{C_3}\]
Answer
161.1k+ views
Hint: The solution for this question will be found by expand the given series and then apply the \[{}^n{C_r}\] properties.
Formula used:In binomial theorem, one of the properties of binomial coefficients is,
\[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\], where
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] is the combination formula, which to be used to find the achievable or possible arrangements. In this formula, \[n\] is the total number of objects in an appropriate group and \[r\] is to be selected from the objects of that group.
Complete step by step solution: Given expression:
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} \]
Now, it is time to expand the series in the given question. Expand the series by putting the values given in the sigma for \[r\], that is from \[r = 1\] to \[r = 6\] and these six values take as the following:
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{50}{C_4} + {}^{56 - 1}{C_3} + {}^{56 - 2}{C_3} + {}^{56 - 3}{C_3} + {}^{56 - 4}{C_3} + {}^{56 - 5}{C_3} + {}^{56 - 6}{C_3}\]
\[ = {}^{50}{C_4} + {}^{55}{C_3} + {}^{54}{C_3} + {}^{53}{C_3} + {}^{52}{C_3}{ + ^{51}}{C_3} + {}^{50}{C_3}\]
\[ = {}^{50}{C_4} + {}^{50}{C_3}{ + ^{51}}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
Now, use the formula for the first two terms of the above series and we can write as
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{51}{C_4}{ + ^{51}}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{52}{C_4} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{53}{C_4} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\\ = {}^{54}{C_4} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{55}{C_4} + {}^{55}{C_3}\\{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{56}{C_4}\]
This is the required value of the given question.
Thus, Option (B) is correct.
Note: The part to be confused by the student is to combine the two terms \[{}^n{C_r} + {}^n{C_{r - 1}}\] . Because, we have used this formula continuously in the further steps as \[r\]is different and there may be confusion between \[4\] and \[3\]. For example, if there is \[{}^n{C_4} + {}^n{C_5}\], we should take it’s answer as \[{}^n{C_5}\] and not as \[{}^n{C_4}\] because \[r - 1 = 4\], \[r = 5\] and we have to take the result term as \[r\] only and not take \[r - 1\]. So, we should not get confused between \[r\] and \[r - 1\]. Simply said, we will consider only the biggest value for \[r\] from the two terms of this formula.
Formula used:In binomial theorem, one of the properties of binomial coefficients is,
\[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\], where
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] is the combination formula, which to be used to find the achievable or possible arrangements. In this formula, \[n\] is the total number of objects in an appropriate group and \[r\] is to be selected from the objects of that group.
Complete step by step solution: Given expression:
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} \]
Now, it is time to expand the series in the given question. Expand the series by putting the values given in the sigma for \[r\], that is from \[r = 1\] to \[r = 6\] and these six values take as the following:
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{50}{C_4} + {}^{56 - 1}{C_3} + {}^{56 - 2}{C_3} + {}^{56 - 3}{C_3} + {}^{56 - 4}{C_3} + {}^{56 - 5}{C_3} + {}^{56 - 6}{C_3}\]
\[ = {}^{50}{C_4} + {}^{55}{C_3} + {}^{54}{C_3} + {}^{53}{C_3} + {}^{52}{C_3}{ + ^{51}}{C_3} + {}^{50}{C_3}\]
\[ = {}^{50}{C_4} + {}^{50}{C_3}{ + ^{51}}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
Now, use the formula for the first two terms of the above series and we can write as
\[{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{51}{C_4}{ + ^{51}}{C_3} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{52}{C_4} + {}^{52}{C_3} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{53}{C_4} + {}^{53}{C_3} + {}^{54}{C_3} + {}^{55}{C_3}\\ = {}^{54}{C_4} + {}^{54}{C_3} + {}^{55}{C_3}\]
\[ = {}^{55}{C_4} + {}^{55}{C_3}\\{}^{50}{C_4} + \sum\limits_{r = 1}^6 {{}^{56 - r}{C_3}} = {}^{56}{C_4}\]
This is the required value of the given question.
Thus, Option (B) is correct.
Note: The part to be confused by the student is to combine the two terms \[{}^n{C_r} + {}^n{C_{r - 1}}\] . Because, we have used this formula continuously in the further steps as \[r\]is different and there may be confusion between \[4\] and \[3\]. For example, if there is \[{}^n{C_4} + {}^n{C_5}\], we should take it’s answer as \[{}^n{C_5}\] and not as \[{}^n{C_4}\] because \[r - 1 = 4\], \[r = 5\] and we have to take the result term as \[r\] only and not take \[r - 1\]. So, we should not get confused between \[r\] and \[r - 1\]. Simply said, we will consider only the biggest value for \[r\] from the two terms of this formula.
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