Question

What is the value of the expression \[{}^n{C_r} + {}^{n - 1}{C_r} + \cdots + {}^r{C_r}\]?
A. \[{}^{n + 1}{C_r}\]
B. \[{}^{n + 1}{C_{r + 1}}\]
C. \[{}^{n + 2}{C_r}\]
D. \[{2^n}\]

Answer 1

Hint: The given expression is the sum of combination. First we will rewrite the given expression. Then we will apply the formula of sum of two combinations in first two terms. In the same we will add all terms of the given expression to get the required answer.



Formula Used:The sum of two combination terms
\[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\]



Complete step by step solution:The given expression is \[{}^n{C_r} + {}^{n - 1}{C_r} + \cdots + {}^r{C_r}\]
Rewrite the given expression:
\[ = {}^r{C_r} + {}^{r + 1}{C_r} + {}^{r + 2}{C_r} + \cdots + {}^{n - 1}{C_r} + {}^n{C_r}\] …..(i)
We know that, \[{}^n{C_n} = 1\].
Therefore \[{}^r{C_r} = {}^{r + 1}{C_{r + 1}}\]
Putting \[{}^r{C_r} = {}^{r + 1}{C_{r + 1}}\] in (i)
\[ = \left( {{}^{r + 1}{C_{r + 1}} + {}^{r + 1}{C_r}} \right) + {}^{r + 2}{C_r} + \cdots + {}^{n - 1}{C_r} + {}^n{C_r}\]
Now applying the formula \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\] in first two terms
\[ = \left( {{}^{r + 2}{C_{r + 1}}} \right) + {}^{r + 2}{C_r} + \cdots + {}^{n - 1}{C_r} + {}^n{C_r}\]
Again applying the formula \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\] in first two terms
\[ = {}^{r + 2}{C_{r + 2}} + \cdots + {}^{n - 1}{C_r} + {}^n{C_r}\]
Similar way
\[ = {}^{n - 1}{C_{r + 1}} + {}^{n - 1}{C_r} + {}^n{C_r}\]
\[ = {}^n{C_{r + 1}} + {}^n{C_r}\]
\[ = {}^{n + 1}{C_{r + 1}}\]



Option ‘B’ is correct

Additional Information:A combination is a combination when r things are chosen from n things and the order of section does not matter.
The combination is represented by \[{}^n{C_r}\].



Note: Students often apply a wrong formula to solve it. They applied \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^n{C_{r + 1}}\] which is an incorrect formula. The correct formula is \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\].