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

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

Explanation

Solution

Here, we will use Pascal’s formula twice which states for any positive natural numbers pp and qq with $$$$:
pCq+pCq−1=p+1Cq{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q} where pCq=p!q!(p−q)!{}^p{C_q} = \dfrac{{p!}}{{q!\left( {p - q} \right)!}}, pCq−1=p!(q−1)!(p−(q−1))!{}^p{C_{q - 1}} = \dfrac{{p!}}{{\left( {q - 1} \right)!\left( {p - \left( {q - 1} \right)} \right)!}} and p+1Cq=(p+1)!(q)!(p+1−q)!{}^{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(n,r)+2C(n,r−1)+C(n,r−2)C\left( {n,r} \right) + 2C\left( {n,r - 1} \right) + C\left( {n,r - 2} \right) in the form of pCq{}^p{C_q} we will get:
C(n,r)+2C(n,r−1)+C(n,r−2)=nCr+2nCr−1+nCr−2C\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 2nCr−12{}^n{C_{r - 1}} into two parts as 2nCr−1=nCr−1+nCr−12{}^n{C_{r - 1}} = {}^n{C_{r - 1}} + {}^n{C_{r - 1}} we get:

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=np = nand $$$$ in pCq+pCq−1=p+1Cq{}^p{C_q} + {}^p{C_{q - 1}} = {}^{p + 1}{C_q} we get:

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 (n+1)!(r−1)(n+1−r)!\dfrac{{\left( {n + 1} \right)!}}{{(r - 1)\left( {n + 1 - r} \right)!}} common and solving for the final result:
⇒nCr+2nCr−1+nCr−2=(n+1)!(r−1)(n+1−r)!(1r+1n−r+2)\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:
⇒(n+1)!(r−1)(n+1−r)!(n+2r(n−r+2))\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 (n+1)!×(n+2)=(n+2)!\left( {n + 1} \right)! \times \left( {n + 2} \right) = \left( {n + 2} \right)!, r×(r−1)!=r!r \times \left( {r - 1} \right)! = r!, and (n−r+1)!×(n−r+2)=(n−r+2)!\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 (n+1)!(r−1)(n+1−r)!(n+2r(n−r+2))\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 nCr{}^n{C_r}, the final result will become n+2Cr{}^{n + 2}{C_r} or C(n+2,r)C\left( {n + 2,r} \right)
So, better to use the pascal method for avoiding the complexity level to calculate the correct answer.