Question

Which of the following is equal to the expression \[{}^n{C_r} + 4 \cdot {}^n{C_{r - 1}} + 6 \cdot {}^n{C_{r - 2}} + 4 \cdot {}^n{C_{r - 3}} + {}^n{C_{r - 4}}\].
A \[{}^{n + 4}{C_r}\]
B \[2 \cdot {}^{n + 4}{C_{r - 1}}\]
C \[4 \cdot {}^n{C_r}\]
D \[11 \cdot {}^n{C_r}\]

Answer 1

Hint: Selection of items from a given collection irrespective of order of section is called combination.

Formula Used: \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\]
Here, n be the total items and r be the selected items.

Complete step by step solution: The given expression is \[{}^n{C_r} + 4 \cdot {}^n{C_{r - 1}} + 6 \cdot {}^n{C_{r - 2}} + 4 \cdot {}^n{C_{r - 3}} + {}^n{C_{r - 4}}\].
First arrange the terms and simplify the expression.
\[{}^n{C_r} + 4 \cdot {}^n{C_{r - 1}} + 6 \cdot {}^n{C_{r - 2}} + 4 \cdot {}^n{C_{r - 3}} + {}^n{C_{r - 4}} = \left( {{}^n{C_r} + {}^n{C_{r - 1}}} \right) + 3 \cdot \left( {{}^n{C_{r - 1}} + {}^n{C_{r - 2}}} \right) + 3 \cdot \left( {{}^n{C_{r - 2}} + {}^n{C_{r - 3}}} \right) + \left( {{}^n{C_{r - 3}} + {}^n{C_{r - 4}}} \right)\]
Now use the concept of \[{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}\] and simplify the expression.
\[ \Rightarrow {}^{n + 1}{C_r} + 3 \cdot {}^{n + 1}{C_{r - 1}} + 3 \cdot {}^{n + 1}{C_{r - 2}} + {}^{n + 1}{C_{r - 3}}\]
Rearrange the terms and write the expression
\[ \Rightarrow \left( {{}^{n + 1}{C_r} + {}^{n + 1}{C_{r - 1}}} \right) + 2\left( {{}^{n + 1}{C_{r - 1}} + {}^{n + 1}{C_{r - 2}}} \right) + \left( {{}^{n + 1}{C_{r - 2}} + {}^{n + 1}{C_{r - 3}}} \right)\]
Simplify the terms as follows.
\[ \Rightarrow \left( {{}^{n + 1}{C_r} + {}^{n + 1}{C_{r - 1}}} \right) + 2\left( {{}^{n + 1}{C_{r - 1}} + {}^{n + 1}{C_{r - 2}}} \right) + \left( {{}^{n + 1}{C_{r - 2}} + {}^{n + 1}{C_{r - 3}}} \right)\]
Now, simplify the expression as follows.
\[ \Rightarrow {}^{n + 2}{C_r} + 2 \cdot {}^{n + 2}{C_{r - 1}} + {}^{n + 2}{C_{r - 2}}\]
Hence, the expression becomes.
\[ \Rightarrow \left( {{}^{n + 2}{C_r} + {}^{n + 2}{C_{r - 1}}} \right) + \left( {{}^{n + 2}{C_{r - 1}} + {}^{n + 2}{C_{r - 2}}} \right)\]
Again, simplify the expression
\[ \Rightarrow {}^{n + 3}{C_r} + {}^{n + 3}{C_{r - 1}}\]
So, the expression is reduced as follows.
\[ \Rightarrow {}^{n + 4}{C_r}\]
The expression \[{}^n{C_r} + 4 \cdot {}^n{C_{r - 1}} + 6 \cdot {}^n{C_{r - 2}} + 4 \cdot {}^n{C_{r - 3}} + {}^n{C_{r - 4}}\] is equal to \[{}^{n + 4}{C_r}\].


Option ‘A’ is correct

Note: The common mistake students make is that any term is missed that gives the wrong answer.