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: Understanding the Question:
The question presents an algebraic equation involving a series of combinatoric binomial coefficients and requires us to solve for the unknown index value $r$.

Step 2: Key Formula or Approach:
We will systematically use Pascal's Identity for combinatorics, which states:
$${^{n}\text{C}_r} + {^{n}\text{C}_{r-1}} = {^{n+1}\text{C}_r}$$

Step 3: Detailed Explanation:
Let's group and simplify the left-hand side of the given equation step-by-step from left to right:
$$\text{LHS} = \left(^{11}\text{C}_4 + {^{11}\text{C}_5}\right) + {^{12}\text{C}_6} + {^{13}\text{C}_7}$$ 1. Apply Pascal's Identity to the first two terms where $n = 11$ and $r = 5$:
$$^{11}\text{C}_4 + {^{11}\text{C}_5} = {^{12}\text{C}_5}$$ Substitute this back into the expression:
$$\text{LHS} = \left(^{12}\text{C}_5 + {^{12}\text{C}_6}\right) + {^{13}\text{C}_7}$$ 2. Apply Pascal's Identity to the next pair where $n = 12$ and $r = 6$:
$$^{12}\text{C}_5 + {^{12}\text{C}_6} = {^{13}\text{C}_6}$$ Substitute this back into our expression:
$$\text{LHS} = {^{13}\text{C}_6} + {^{13}\text{C}_7}$$ 3. Apply Pascal's Identity to the final pair where $n = 13$ and $r = 7$:
$${^{13}\text{C}_6} + {^{13}\text{C}_7} = {^{14}\text{C}_7}$$ Now equate our simplified LHS to the given RHS from the problem statement:
$${^{14}\text{C}_7} = {^{14}\text{C}_r} \implies r = 7$$ This matches option (C).

Step 4: Final Answer:
The value of $r$ is 7, which corresponds to option (C).