Question

Find the value of \[C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right)\]:

Answer 1

Hint: Here, we will use Pascal’s formula twice which states for any positive natural numbers \[p\] and \[q\] with \[\]:
 \[{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}\] 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 \[{}^{p + 1}{C_q} = \dfrac{{\left( {p + 1} \right)!}}{{\left( q \right)!\left( {p + 1 - q} \right)!}}\].

Complete step-by-step answer:
Step 1: By writing the expression \[C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right)\] in the form of \[{}^p{C_q}\] we will get:
\[C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right) = {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}}\]….…… (1)
Step 2: Now, by simplifying the expression (1) and breaking the term \[2{}^n{C_{r - 1}}\] into two parts as \[2{}^n{C_{r - 1}} = {}^n{C_{r - 1}} + {}^n{C_{r - 1}}\] we get:
\[\]………………………. (2)
By solving the above expression (2) in two separate steps by using Pascal’s rule which states that for whole numbers \[p\]and\[q\] where \[p \geqslant q\]:
\[\]
Step 3: Using Pascal’s rule for the first two terms in the expression (2) by putting \[p = n\]and \[\] in \[{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}\] we get:
\[\] ………………….. (3)
Similarly, by using Pascal’s rule for the last two terms in the expression (2) by putting \[p = n\]and \[q = r - 1\] in \[{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}\] we get:
\[ \Rightarrow {}^n{C_{r - 1}} + {}^n{C_{r - 2}} = {}^{n + 1}{C_{r - 1}}{\text{ }}\left( {\because {}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}} \right)\]…………………. (4)
So, the final term \[{}^n{C_r} + {}^n{C_{r - 1}} + {}^n{C_{r - 1}} + {}^n{C_{r - 2}}\] will be written as the addition of expression (3) and (4):
\[ \Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} + {}^n{C_{r - 1}} + {}^n{C_{r - 2}} = {}^{n + 1}{C_r} + {}^{n + 1}{C_{r - 1}}\]
Therefore, from equation (2) we get
\[ \Rightarrow {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}}{\text{ }} = {}^{n + 1}{C_r} + {}^{n + 1}{C_{r - 1}}\] …………………. (5)
Step 4: Now, in the above expression (5), by applying the same pascal rule by putting \[p = n + 1\]and \[q = r\] in \[{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}\] we get:
\[ \Rightarrow {}^{n + 1}{C_r} + {}^{n + 1}{C_{r - 1}} = {}^{(n + 1) + 1}{C_r} = {}^{n + 2}{C_r}\left( {\because {}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q}} \right)\]
We can also write the above answer \[{}^{n + 2}{C_r}\] as below:
\[C\left( {n + 2,r} \right)\]

Therefore, \[C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right) = C\left( {n + 2,r} \right)\].

Note: Students can also try to go for expanding and solving the above term \[C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right)\]by expanding the terms \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], \[2{}^n{C_{r - 1}} = 2\left( {\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)!}}} \right)\]and \[{}^n{C_{r - 2}} = \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - \left( {r - 2} \right)} \right)!}}\]:
\[{}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} + 2\left( {\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}} \right) + \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!}}\]……………… (1)
By expanding the term \[2\left( {\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}} \right) = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}\]in the above equation (1):
\[ \Rightarrow {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} + \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!}}\]
Now we will consider the first two and last two terms:
\[ \Rightarrow {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}} = \left( {\dfrac{{n!}}{{r!\left( {n - r} \right)!}} + \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}} \right) + \left( {\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} + \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!}}} \right)\]
For the left and right bracket, by taking \[\dfrac{{n!}}{{(r - 1)\left( {n - r} \right)!}}\], \[\dfrac{{n!}}{{(r - 1)\left( {n - r + 1} \right)!}}\] common respectively we have:
\[ \Rightarrow {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 1}}} \right) + \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{1}{{r - 1}} + \dfrac{1}{{n - r + 2}}} \right)\] …… (2)
We will solve the above expression (2) in two separate parts:
For the first part i.e. \[\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 1}}} \right)\], by multiplying and dividing the two fractions inside the bracket using common factors we ensure that denominators are the same:
\[\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 1}}} \right) = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{1 \cdot \left( {n - r + 1} \right)}}{{r \cdot \left( {n - r + 1} \right)}} + \dfrac{{1 \cdot r}}{{r \cdot \left( {n - r + 1} \right)}}} \right)\]… (3)
Now, in the above equation (3) by multiplying the factors and adding the two fractions we get:
\[ \Rightarrow \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{1 \cdot \left( {n - r + 1} \right)}}{{r \cdot \left( {n - r + 1} \right)}} + \dfrac{{1 \cdot r}}{{r \cdot \left( {n - r + 1} \right)}}} \right) = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{n - r + 1 + r}}{{r \cdot \left( {n - r + 1} \right)}}} \right)\]
\[ \Rightarrow \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 1}}} \right) = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}}\left( {\dfrac{{n + 1}}{{r \cdot \left( {n - r + 1} \right)}}} \right)\]…. (4)
For the last two parts of equation (2) i.e. \[\dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{1}{{r - 1}} + \dfrac{1}{{n - r + 2}}} \right)\], by multiplying and dividing the two fractions inside the bracket using common factors, we ensure that denominators are the same:
\[\dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{1}{{r - 1}} + \dfrac{1}{{n - r + 2}}} \right) = \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{{1 \cdot \left( {n - r + 2} \right)}}{{\left( {r - 1} \right) \cdot \left( {n - r + 2} \right)}} + \dfrac{{1 \cdot \left( {r - 1} \right)}}{{\left( {r - 1} \right) \cdot \left( {n - r + 2} \right)}}} \right)\]….. (5)
Now, in the above equation (5) by multiplying the factors and adding the two fractions we get:
\[\dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{{1 \cdot \left( {n - r + 2} \right)}}{{\left( {r - 1} \right) \cdot \left( {n - r + 2} \right)}} + \dfrac{{1 \cdot \left( {r - 1} \right)}}{{\left( {r - 1} \right) \cdot \left( {n - r + 2} \right)}}} \right) = \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{{n - r + 2 + r - 1}}{{\left( {r - 1} \right) \cdot \left( {n - r + 2} \right)}}} \right)\] \[\] … (6)
Now, for equation (4), 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 equation (4):
\[\] …………………… (7)
For equation (6), we can write \[n! \times \left( {n + 1} \right) = \left( {n + 1} \right)!\], \[\left( {r - 1} \right) \times \left( {r - 2} \right)! = \left( {r - 1} \right)!\] and \[\left( {n - r + 1} \right)! \times \left( {n - r + 2} \right) = \left( {n - r + 2} \right)!\]. Substituting these values in the above equation (6):
\[ \Rightarrow \dfrac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}\left( {\dfrac{1}{{r - 1}} + \dfrac{1}{{n - r + 2}}} \right) = \dfrac{{\left( {n + 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - r + 2} \right)!}}\] ………………….. (8)
Finally combining both equations (7) and (8) and comparing with equation (2):
\[\]
Taking \[\dfrac{{\left( {n + 1} \right)!}}{{(r - 1)\left( {n + 1 - r} \right)!}}\] common and solving for the final result:
\[ \Rightarrow {}^n{C_r} + 2{}^n{C_{r - 1}} + {}^n{C_{r - 2}} = \dfrac{{\left( {n + 1} \right)!}}{{\left( {r - 1} \right)\left( {n + 1 - r} \right)!}}\left( {\dfrac{1}{r} + \dfrac{1}{{n - r + 2}}} \right)\]
Now by repeating the same steps as in equation (5) and (6), we will get:
\[ \Rightarrow \dfrac{{\left( {n + 1} \right)!}}{{(r - 1)\left( {n + 1 - r} \right)!}}\left( {\dfrac{{n + 2}}{{r\left( {n - r + 2} \right)}}} \right)\] …………….. (9)
Now, for the above expression (9), we can write \[\left( {n + 1} \right)! \times \left( {n + 2} \right) = \left( {n + 2} \right)!\], \[r \times \left( {r - 1} \right)! = r!\], and \[\left( {n - r + 1} \right)! \times \left( {n - r + 2} \right) = \left( {n - r + 2} \right)!\], substituting these values in the above equation (9):
By comparing the above term \[\dfrac{{\left( {n + 1} \right)!}}{{(r - 1)\left( {n + 1 - r} \right)!}}\left( {\dfrac{{n + 2}}{{r\left( {n - r + 2} \right)}}} \right)\] in the form of \[{}^n{C_r}\], the final result will become \[{}^{n + 2}{C_r}\] or \[C\left( {n + 2,r} \right)\]
So, better to use the pascal method for avoiding the complexity level to calculate the correct answer.