Question

Find the value of \[{}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} \].
A. \[{}^{18}{C_3}\]
B. \[{}^{18}{C_4}\]
C. \[{}^{14}{C_7}\]
D. None of these

Answer 1

Hint: First, simplify the summation by substituting the values of \[j\]. Then, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\] and solve the equation. Again, use this property and solve the expression to get the required answer.



Formula Used:\[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\]



Complete step by step solution:The given expression is \[{}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} \].
Simplify the expression by substituting the values of \[j\].
\[{}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_4} + {}^{18 - 1}{C_3} + {}^{18 - 2}{C_3} + {}^{18 - 3}{C_3} + {}^{18 - 4}{C_3}\]
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_4} + {}^{17}{C_3} + {}^{16}{C_3} + {}^{15}{C_3} + {}^{14}{C_3}\]
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_3} + {}^{14}{C_4} + {}^{15}{C_3} + {}^{16}{C_3} + {}^{17}{C_3}\]
Apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
We get,
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{15}{C_4} + {}^{15}{C_3} + {}^{16}{C_3} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{16}{C_4} + {}^{16}{C_3} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{17}{C_4} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{18}{C_4}\]



Option ‘B’ is correct



Note: The given expression is \[{}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} \].
Simplify the expression by substituting the values of \[j\].
\[{}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_4} + {}^{18 - 1}{C_3} + {}^{18 - 2}{C_3} + {}^{18 - 3}{C_3} + {}^{18 - 4}{C_3}\]
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_4} + {}^{17}{C_3} + {}^{16}{C_3} + {}^{15}{C_3} + {}^{14}{C_3}\]
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{14}{C_3} + {}^{14}{C_4} + {}^{15}{C_3} + {}^{16}{C_3} + {}^{17}{C_3}\]
Apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
We get,
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{15}{C_4} + {}^{15}{C_3} + {}^{16}{C_3} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{16}{C_4} + {}^{16}{C_3} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{17}{C_4} + {}^{17}{C_3}\]
Again, apply the combination property \[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}\].
\[ \Rightarrow {}^{14}{C_4} + \sum\limits_{j = 1}^4 {{}^{18 - j}{C_3}} = {}^{18}{C_4}\]