Question

Find the value of ${}^n{C_r} + {}^n{C_{r - 1}}$:

Answer 1

Here we will use the combination formula for expanding the terms which state that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $n$ is the total number of items in the set and $r$ is the selected ones from that set.

Complete step-by-step answer:
Step 1: By expanding the term ${}^n{C_r} + {}^n{C_{r - 1}}$ as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ , ${}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)!}}$and writing the term as below:
${}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)!}}$
Taking $\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}$ common in RHS side of ${}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)!}}$, we get:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 1}}} \right)$ ………… (1)
Step 2: Now in the above equation (1), by multiplying and dividing the two fractions inside the bracket using common factors. We ensure that denominators are the same:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{1 \cdot \left( {n - r + 1} \right)}}{{r\left( {n - r + 1} \right)}} + \dfrac{{1 \cdot r}}{{r\left( {n - r + 1} \right)}}} \right)$………… (2)
Now, in the above equation (2) by multiplying the factors and adding the two fractions we get:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{n - r + 1 + r}}{{r\left( {n - r + 1} \right)}}} \right)$
By simplifying inside the brackets, we get:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{n + 1}}{{r\left( {n - r + 1} \right)}}} \right)$……….. (3)
Step 3: Now, we can write $n! \times \left( {n + 1} \right) = \left( {n + 1} \right)!$, $r \times \left( {r - 1} \right)! = r!$ and $\left( {n - r} \right)! \times \left( {n - r + 1} \right) = \left( {n - r + 1} \right)!$ substituting these values in the above expression (3):
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = \dfrac{{\left( {n + 1} \right)!}}{{r!\left( {n - r + 1} \right)!}}$
Step 4: By comparing the above term $\dfrac{{\left( {n + 1} \right)!}}{{r!\left( {n - r + 1} \right)!}}$ from the combination formula ${}^p{C_q} = \dfrac{{p!}}{{q!\left( {p - q} \right)!}}$, the final result will be:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$

So, the answer is ${}^{n + 1}{C_r}$.

Note: Students can also try to solve these types of the problem by using the pascal formula which states for any positive natural numbers $p$ and $q$ with $p \geqslant q$:
${}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}$ where ${}^p{C_q} = \dfrac{{p!}}{{q!\left( {p - q} \right)!}}$, ${}^p{C_{q - 1}} = \dfrac{{p!}}{{\left( {q - 1} \right)!\left( {p - \left( {q - 1} \right)} \right)!}}$ and ${}^{p + 1}{C_q} = \dfrac{{\left( {p + 1} \right)!}}{{\left( q \right)!\left( {p + 1 - q} \right)!}}$
Step 1: We need to find the value of ${}^n{C_r} + {}^n{C_{r - 1}}$. By applying combination rule here which states that:
To calculate the total number of outcomes of an event where $n$represents the total number of items in the set and $r$represents the number of selected items being chosen at a time and $C$ states for combination:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Step 2: Now, for finding the value of the term ${}^n{C_r} + {}^n{C_{r - 1}}$ , we will use the pascal rule which states that for $p \geqslant q$ and $p$ & $q$ are positive natural numbers:
$\Rightarrow {}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}$
Step 3: By comparing the term ${}^n{C_r} + {}^n{C_{r - 1}}$ with the pascal formula ${}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}$ , we can write the expression as below:
$\Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}{\text{ (}}\because {\text{by using pascal's rule)}}$
Where, $p = n$ and $q = r$.
Students should not confuse between the permutation and combination formula. There is a major difference between the formula for both terms:
Permutation formula: ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, Combination formula: ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .