Question

Find the value of\[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]if \[0 \le r \le n - 1\].
A. \[\left( {\begin{array}{*{20}{c}}n\\{r - 1}\end{array}} \right)\]
B. \[\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right)\]
C. \[\left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]
D. \[\left( {\begin{array}{*{20}{c}}{n + 1}\\{r + 1}\end{array}} \right)\]

Answer 1

Hint: In the given question, we need to find the value of \[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]. For this, we will use the property of combination rule and the given condition such as \[0 \le r \le n - 1\] to get the value of \[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\].

Formula used: The following formula used for solving the given question.
The formula of combination is given by, \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, \[n\] is the total number of things and \[r\] is the number of things that need to be selected from total things.

Complete step by step solution: We know that \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, \[n\] is the total number of things and \[r\] is the number of things that need to be selected from total things.
Now, we need to find the value of \[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]
Thus, we can say that
. \[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right) = {}^n{C_{n - r}} + {}^n{C_{r + 1}}\]
But according to property of combination, we can say that
 \[{}^n{C_x} = {}^n{C_y} \Rightarrow n = x + y{\rm{ or }}x = y\]
Thus, we will get
\[n - r = r + 1\]
This gives
\[n = 2r + 1\]
Thus, we get
 \[n = 2r + 1 \Rightarrow {}^n{C_r} + {}^n{C_{r + 1}}\]
By simplifying, we get
\[{}^{n + 1}{C_{r + 1}} = \left( {\begin{array}{*{20}{c}}{n + 1}\\{r + 1}\end{array}} \right)\]
Hence, The value of\[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]if \[0 \le r \le n - 1\], is \[\left( {\begin{array}{*{20}{c}}{n + 1}\\{r + 1}\end{array}} \right)\].

Thus, Option (A) is correct.

Additional Information: Combinations are ways to choose elements from a group in mathematics in which the order of the selection is irrelevant. Permutations and combinations can be jumbled up. The sequence in which the chosen components are chosen is crucial in permutations, too. Combinatorics is the study of combinations; however, mathematics as well as finance are two other fields that use combinations.

Note: Many students make mistakes in calculation as well as writing the property of combination rule. This is the only way through which we can solve the example in the simplest way. Also, it is necessary to find the appropriate value of \[\left( {\begin{array}{*{20}{c}}n\\{n - r}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}n\\{r + 1}\end{array}} \right)\]such that it holds the condition such as \[0 \le r \le n - 1\].