Three letters can be posted in five letterboxes in $3^5$ ways.

Three letters can be posted in five letterboxes in 3⁵ ways.

NCERT Exemplar · Q52

3^5

Odisha ChseShort· 1mImportance★★★★★

91% · 118/130 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Combinations Probability

Combinations Probability: Counting When Order Doesn't Matter

Imagine you are picking a team of 3 students from a class of 10. You do not care who is chosen first, second, or third — you only care which 3 students end up on the team. That is a combination: a selection where order does not matter.

If you now ask for the chance that one particular set of 3 students (say your three best friends) is the one chosen, you are doing combinations probability: probability where the favourable and total outcomes are both counted using combinations.

The Core Intuition

When every possible selection is equally likely (like drawing names from a hat), the probability of an event is the familiar ratio:

P(E) = \frac{\text{Number of favourable selections}}{\text{Total number of possible selections}}

This is the same "favourable over total" idea from basic probability — the only new part is that we count combinations, not arrangements, because order is irrelevant.

Note

The key difference from permutations: the group $\{$ Alice, Bob, Charlie $\}$ is the same selection as $\{$ Charlie, Bob, Alice $\}$ . Swapping the order of chosen items does not create a new outcome.

The Precise Statement

Let $n$ be the total number of distinct objects and let $r$ be how many you choose (without replacement). The number of ways to choose $r$ objects from $n$ is:

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

read as " $n$ choose $r$ ". If a selection of $r$ objects is made at random and every combination is equally likely, then the probability of an event $E$ is:

P(E) = \frac{\text{number of combinations in } E}{\dbinom{n}{r}}

A Worked Example

Problem: A bag has 5 red marbles and 3 blue marbles. You draw 3 marbles at random (without looking). What is the probability of getting exactly 2 red marbles?

Step 1 — Total outcomes. Choosing 3 marbles from 8:

\binom{8}{3} = \frac{8!}{3!\,5!} = 56

Step 2 — Favourable outcomes. You need exactly 2 red (from 5) and 1 blue (from 3):

\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30

Step 3 — Probability.

P(\text{exactly 2 red}) = \frac{30}{56} = \frac{15}{28}

Tip

For "exactly $k$ of one type", multiply (ways to choose $k$ from that type) by (ways to choose the rest from the others), then divide by the total number of combinations.

Combinations vs. Permutations

| Situation | Use |

|-----------|-----| …

Start your 14-day free trial to unlock the full solution →

The statement is false. The number of ways to post 3 distinct letters into 5 letterboxes is $5^3$ , not $3^5$ . The key is to decide which box each letter goes into — each letter has 5 choices, giving $5 \times 5 \times 5 = 125$ ways.

Why the given statement is wrong

The confusion here is a classic one: who chooses what? When you post letters, each letter independently picks a letterbox. That means the letter is the active agent, and the letterbox is the destination. So for each of the 3 letters, there are 5 possible boxes — not the other way around.

Let’s break it down.

1. Identify the "objects" and the "containers"

We have:

Objects: 3 distinct letters (say L₁, L₂, L₃)
Containers: 5 distinct letterboxes (say B₁, B₂, B₃, B₄, B₅)

The act of posting means: for each letter, choose one box to drop it into. Multiple letters can go into the same box — there’s no restriction.

2. Count the choices per letter

Letter L₁ can go into any of the 5 boxes → 5 choices.

Letter L₂ can also go into any of the 5 boxes → 5 choices.

Letter L₃ similarly → 5 choices.

Since the choices for different letters are independent, we multiply:

5 \times 5 \times 5 = 5^3 = 125

That’s the total number of ways.

3. Why $3^5$ would be wrong

$3^5$ would mean: for each of the 5 boxes, you choose one of the 3 letters to put into it. That would be the count if you were distributing boxes among letters — for example, if each box could receive at most one letter, or if you were assigning a "letter" to each box. But here, each letter must go somewhere, and boxes can hold many letters. The roles are reversed.

Watch out

A common mistake is to swap the base and exponent. Remember: the number of ways to assign $n$ distinct items to $r$ distinct bins (with no limit on bin capacity) is $r^n$ , not $n^r$ . The exponent is the number of items, the base is the number of bins.

4. A quick check with small numbers …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ51 — There are 12 points in a plane of which 5 points are collinear, then t…→Next questionQ53 — In the permutations of n things, r taken together, the number of permu…

Q.Three letters can be posted in five letterboxes in 353^535 ways.

Why the given statement is wrong

1. Identify the "objects" and the "containers"

2. Count the choices per letter

3. Why 353^535 would be wrong

4. A quick check with small numbers …

🎓Unlock everything free for 14 days

Q.Three letters can be posted in five letterboxes in $3^5$ ways.

3. Why $3^5$ would be wrong

Unlock everything free for 14 days