If ⁿC₄, ⁿC₅ and ⁿC₆ are in A.P., then n can be: (1) 14 (2) 11 (3) 9 (4) 12

Q: Why can't we just use the property $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$?

That property is excellent for simplifying sums of binomial coefficients, but here we have an A.P. condition, which is a multiplicative relationship ($2b = a+c$). The property you mentioned doesn't directly help with the A.P. structure. Stick to the definition of the combination formula to avoid getting lost in identities that don't apply here.

Q: How do I know which factorial to expand?

Always look for the smallest factorial in the denominator. For example, between $4!$, $5!$, and $6!$, the smallest is $4!$. Express $5!$ as $5 \times 4!$ and $6!$ as $6 \times 5 \times 4!$. This creates a common factor that you can cancel across the entire equation. The same logic applies to the $(n-r)!$ terms.

Q: I got two answers ($n=7$ and $n=14$), but only one is in the options. Is $n=7$ wrong?

In JEE Advanced, sometimes multiple values are mathematically valid, but only one is provided in the options. Both $n=7$ and $n=14$ satisfy the A.P. condition. Since $14$ is the only one listed, it is the correct choice. Always check your derived values against the provided options.

Question

If $^nC_4, ^nC_5$ and $^nC_6$ are in A.P., then $n$ can be:

Options:

(A) $14$

(B) $11$

(C) $9$

(D) $12$

Sigma Percentile · Accepted Answer

### The Dance of Binomial Coefficients: A Journey into A.P.

Welcome, future engineer. Today, we are going to unravel a problem that sits at the beautiful intersection of sequences and combinatorics. It is a classic JEE Advanced problem that tests not just your knowledge of formulas, but your ability to navigate algebraic complexity with grace.

We are looking at three binomial coefficients: $\binom{n}{4}$, $\binom{n}{5}$, and $\binom{n}{6}$, and we are told they exist in an Arithmetic Progression (A.P.). Let us dive in.

### Phase 1: The Bridge of A.P.

Before we touch a single factorial, we must respect the structure of the problem. If we have three terms $a, b, c$ in an arithmetic progression, the middle term $b$ is the arithmetic mean of the other two.

Mathematically, this is expressed as $2b = a + c$. For our specific problem, this translates to the elegant equation:

$$2 	imes \binom{n}{5} = \binom{n}{4} + \binom{n}{6}$$

This is the bridge that connects the world of sequences to the world of combinations.

### Phase 2: The Factorial Jungle

Now, we enter the forest of factorials. The definition of a combination is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. When we substitute this into our equation, it looks intimidating:

$$2 	imes \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$$

I know, it looks like a mess of symbols. But take a deep breath. Look at the numerators; every single term contains $n!$.

This is the universe giving us a gift! We can divide the entire equation by $n!$, effectively vanishing it from existence. We are left with:

$$\frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}$$

### Phase 3: The Great Simplification

This is where the true artistry of algebra comes in. We need to clear these denominators by expressing the larger factorials in terms of the smaller ones.

For the constants, we note that $5! = 5 	imes 4!$ and $6! = 6 	imes 5 	imes 4! = 30 	imes 4!$. For the variable terms, we observe that $(n-6)!$ is the smallest.

We write $(n-5)! = (n-5) 	imes (n-6)!$ and $(n-4)! = (n-4) 	imes (n-5) 	imes (n-6)!$. When we substitute these back, the common terms $4!$ and $(n-6)!$ appear in every denominator.

Multiplying the entire equation by $4!(n-6)!$ allows us to banish them. We are left with a clean, manageable rational equation:

$$\frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$$

### Phase 4: The Quadratic Climax

We are in the home stretch. To clear the fractions, we multiply by the least common multiple, $30(n-4)(n-5)$. This transforms our rational equation into a standard quadratic:

$$12(n-4) = 30 + (n-4)(n-5)$$

Expanding this, we get $12n - 48 = 30 + n^2 - 9n + 20$. Rearranging everything to one side, we arrive at the beautiful quadratic equation:

$$n^2 - 21n + 98 = 0$$

Factoring this, we find $(n-14)(n-7) = 0$. This gives us two potential values for $n$: $14$ and $7$.

Since the binomial coefficients $\binom{n}{4}, \binom{n}{5}, \binom{n}{6}$ require $n \ge 6$, both are mathematically valid, but **$n=14$** and **$n=7$** are the solutions. You have successfully navigated the complexity and arrived at the truth. Keep this confidence; it is the hallmark of a true problem solver.

Animated Solution for Mathematics - Permutations and Combinations: If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then $n$ can be:

Select Answer:

Visualized Solution

Given Condition

Condition for Arithmetic Progression

Applying the A.P. Condition

The Combination Formula

Expanding the Equation

Eliminating Common Factors

Breaking Down Factorials

Breaking Down Variable Factorials

Substituting Expanded Factorials

Canceling Common Denominator Terms

Clearing the Fractions

Expanding the Terms

Forming the Quadratic Equation

Solving the Quadratic Equation

Final Answer Selection

The Sigma Insight: Combinations and Selection

The Dance of Binomial Coefficients

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If nC4​,nC5​ and nC6​ are in A.P., then n can be: