If $^nC_4, {}^nC_5$ and $^nC_6$ are in AP, find $n$

> [!TLDR] Since ${}^nC_4,{}^nC_5,{}^nC_6$ are in AP, $2\,{}^nC_5={}^nC_4+{}^nC_6$; solving the resulting quadratic gives $n=7$ or $n=14$. > [!FORMULA] For three terms $a,b,c$ in AP: $2b=a+c$. Also $\dfrac{{}^nC_{r-1}}{{}^nC_r}=\dfrac{r}{n-r+1}$ and $\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$, where $n$ = total items, $r$ = items chosen. 1. AP condition: $2\,{}^nC_5 = {}^nC_4 + {}^nC_6$. 2. Divide throughout by ${}^nC_5$: $2 = \dfrac{{}^nC_4}{{}^nC_5} + \dfrac{{}^nC_6}{{}^nC_5}$. 3. Use $\dfrac{{}^nC_4}{{}^nC_5}=\dfrac{5}{n-4}$ and $\dfrac{{}^nC_6}{{}^nC_5}=\dfrac{n-5}{6}$, so $2=\dfrac{5}{n-4}+\dfrac{n-5}{6}$. 4. Multiply both sides by $6(n-4)$: $12(n-4) = 30 + (n-5)(n-4)$. 5. Expand the right side: $12n-48 = 30 + n^2-9n+20 = n^2-9n+50$. 6. Rearrange into a quadratic: $n^2-21n+98=0$. …

If ⁿC₄, ⁿC₅ and ⁿC₆ are in AP, find n

Worked Examples · Example 43

^nC_4, {}^nC_5

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Concept understanding — Combination

Combination: Choosing Without Ordering

Imagine you have three friends: A, B, and C. You need to pick two of them to form a committee. How many different committees can you make?

You could pick A and B, or A and C, or B and C. That's three committees. Notice something crucial: the committee {A, B} is exactly the same as {B, A}. Who gets picked first doesn't matter — only who is on the committee matters. This is the core idea of a combination: a selection where the order of selection does not matter.

The Intuition: "How many groups?"

A combination answers the question: "From a set of $n$ distinct items, how many ways can I choose $r$ of them, ignoring the order in which I pick them?"

Think of it as forming a team, a committee, a hand of cards, or a set of ingredients for a recipe. In all these cases, swapping two members around doesn't give you a new team — it's still the same group.

The Precise Statement

A combination is a selection of $r$ objects from a set of $n$ distinct objects, where the order of selection does not matter. The number of such combinations is denoted by:

\binom{n}{r} \quad \text{or} \quad ^nC_r

and is given by the formula:

\binom{n}{r} = \frac{n!}{r! \, (n-r)!}

where $n!$ (read "n factorial") means $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ , and $0! = 1$ by definition.

Why the $r!$ in the denominator?

This is the key to understanding combinations. If order did matter, the number of ways to pick $r$ objects from $n$ would be $n \times (n-1) \times \cdots \times (n-r+1)$ , which is written as $\frac{n!}{(n-r)!}$ . That's called a permutation.

But in a combination, every group of $r$ objects can be rearranged in $r!$ different orders. Since we don't care about order, we divide by $r!$ to "collapse" all those orderings into one single group.

So:

\binom{n}{r} = \frac{\text{number of ordered selections}}{\text{number of ways to order each selection}} = \frac{n!}{(n-r)!} \div r! = \frac{n!}{r! \, (n-r)!}

A Quick Example

From a class of 10 students, how many ways can you choose a team of 3?

Here $n=10$ , $r=3$ .

\binom{10}{3} = \frac{10!}{3! \, 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120

So there are 120 different teams possible. …

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Since ${}^nC_4,{}^nC_5,{}^nC_6$ are in AP, $2\,{}^nC_5={}^nC_4+{}^nC_6$ ; solving the resulting quadratic gives $n=7$ or $n=14$ .

a,b,c

AP condition: $2\,{}^nC_5 = {}^nC_4 + {}^nC_6$ .
Divide throughout by ${}^nC_5$ : $2 = \dfrac{{}^nC_4}{{}^nC_5} + \dfrac{{}^nC_6}{{}^nC_5}$ .
Use $\dfrac{{}^nC_4}{{}^nC_5}=\dfrac{5}{n-4}$ and $\dfrac{{}^nC_6}{{}^nC_5}=\dfrac{n-5}{6}$ , so $2=\dfrac{5}{n-4}+\dfrac{n-5}{6}$ .
Multiply both sides by $6(n-4)$ : $12(n-4) = 30 + (n-5)(n-4)$ .
Expand the right side: $12n-48 = 30 + n^2-9n+20 = n^2-9n+50$ .
Rearrange into a quadratic: $n^2-21n+98=0$ . …

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Q.If nC4,nC5^nC_4, {}^nC_5nC4​,nC5​ and nC6^nC_6nC6​ are in AP, find nnn

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Q.If $^nC_4, {}^nC_5$ and $^nC_6$ are in AP, find $n$

Unlock everything free for 14 days