Question

If $^{11}\text{C}_4 + {^{11}\text{C}_5} + {^{12}\text{C}_6} + {^{13}\text{C}_7} = {^{14}\text{C}_r}$, then the value of $r$ is

• A. 11
• B. 14
• C. 7
• D. 3

Hint:

Pascal's identity works like a cascading domino rally! As long as the lower indices are sequential ($4$ and $5$) and top elements match ($11$ and $11$), they combine and step up by $1$. Watch the top elements cascade: $11 \rightarrow 12 \rightarrow 13 \rightarrow 14$, pulling the highest lower index along with it!

Answer 1

The Correct Option is C

Step 1: State the tool.
We repeatedly use Pascal's identity $^{n}C_{r-1}+{}^{n}C_{r}={}^{n+1}C_{r}$ to combine adjacent terms.
Step 2: Combine the first pair.
$^{11}C_4+{}^{11}C_5={}^{12}C_5$ (here $n=11$, lower indices 4 and 5).
Step 3: Rewrite the sum.
The expression becomes $^{12}C_5+{}^{12}C_6+{}^{13}C_7$.
Step 4: Combine the next pair.
$^{12}C_5+{}^{12}C_6={}^{13}C_6$, so we now have $^{13}C_6+{}^{13}C_7$.
Step 5: Combine the last pair.
$^{13}C_6+{}^{13}C_7={}^{14}C_7$.
Step 6: Match with the right side.
So the left side equals $^{14}C_7$, and since it equals $^{14}C_r$, we get $r=7$, matching option (3).
\[ \boxed{r=7} \]