Question

Find the value of \[{}^n{C_r} + {}^n{C_{r - 1}}\]:

Answer 1

Hint: 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: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] .