Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a

NCERT Exemplar · Q2

Q. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks? [Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1.]

Odisha ChseShort· 3mImportance★★★★★

56% · 52/93 Questions

✓ Free question

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
Order does not matter (team, hand of cards, lottery numbers)	Combinations
Order does matter (password, race positions, seating)	Permutations

If unsure, ask: "Would swapping two selected items change the outcome?" If no, it is combinations.

Watch out

Do not use plain combinations when items are chosen with replacement (like drawing a card and putting it back). Combinations assume selection without repetition; with replacement the counting changes.

The Big Picture

Combinations probability is just careful counting when order is irrelevant:

Count the total equally likely selections using $\binom{n}{r}$ .
Count how many of those match your event.
Divide.

Combinations Probability links the NCERT Class 11 Mathematics chapters on Permutations & Combinations and Probability, matching searches like "probability using combinations formula class 11" or "probability important questions class 11 maths". This $\binom{n}{r}$ -based approach to counting favourable outcomes is a staple technique in CBSE board and JEE Main problems.

The problem asks for the probability that a married couple among six employees will not be assigned adjacent desks. It's easier to calculate the complementary probability: first find the probability that they are assigned adjacent desks, and then subtract this from 1. The final probability is $\boxed{\frac{2}{3}}$ .

This problem involves calculating probabilities related to arrangements, which falls under permutations. When dealing with "not" scenarios (like non-adjacent), it's often simpler to use the concept of complementary probability. The probability of an event happening is $P(E)$ , and the probability of it not happening is $P(E')$ . These are related by the formula $P(E') = 1 - P(E)$ . Here, calculating the probability that the couple is adjacent is more straightforward than directly calculating the probability that they are not adjacent.

Let's break down the solution step-by-step.

Determine the total number of possible arrangements. We have 6 employees and 6 distinct desks arranged in a row. Each employee can be assigned to exactly one desk, and each desk will have exactly one employee. This is a problem of arranging 6 distinct items in 6 distinct positions. The total number of ways to assign these employees to the desks is given by the number of permutations of 6 items, which is $6!$ .

\text{Total arrangements} = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720

Determine the number of arrangements where the married couple is adjacent.

Let the married couple be $M_1$ and $M_2$ . For them to be adjacent, they must occupy two desks next to each other. We can treat this couple as a single "block" or unit.
- Step 2a: Treat the couple as one unit. If $M_1$ and $M_2$ must sit together, we can consider them as a single entity $(M_1 M_2)$ . Now, instead of 6 individual employees, we effectively have 5 entities to arrange: the couple-unit $(M_1 M_2)$ and the remaining 4 individual employees ( $E_3, E_4, E_5, E_6$ ). The number of ways to arrange these 5 entities is $5!$ .

\text{Arrangements of 5 entities} = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

*   **Step 2b: Consider the internal arrangement of the couple.**
    Within the $(M_1 M_2)$ unit, the couple can be arranged in two ways: $M_1$ on the left and $M_2$ on the right, or $M_2$ on the left and $M_1$ on the right. This means there are $2!$ ways for the couple to arrange themselves within their adjacent desks.

\text{Internal arrangements of couple} = 2! = 2 \times 1 = 2

*   **Step 2c: Calculate total adjacent arrangements.**
    To find the total number of arrangements where the couple is adjacent, we multiply the number of ways to arrange the 5 entities by the number of ways the couple can arrange themselves internally.

\text{Arrangements with couple adjacent} = 5! \times 2! = 120 \times 2 = 240

Calculate the probability that the married couple is adjacent. This probability is the ratio of the number of arrangements where the couple is adjacent to the total number of possible arrangements.

P(\text{couple adjacent}) = \frac{\text{Arrangements with couple adjacent}}{\text{Total arrangements}} = \frac{240}{720}

We can simplify this fraction:

P(\text{couple adjacent}) = \frac{240}{720} = \frac{24}{72} = \frac{1}{3}

> [!TIP]
> A quicker way to simplify $\frac{5! \times 2!}{6!}$ is to expand $6!$ as $6 \times 5!$:
> $$ \frac{5! \times 2!}{6!} = \frac{5! \times 2}{6 \times 5!} = \frac{2}{6} = \frac{1}{3} $$

4. Calculate the probability that the married couple will have nonadjacent desks.

As per the hint and the concept of complementary probability, the probability that the couple has nonadjacent desks is $1$ minus the probability that they have adjacent desks.

P(\text{couple nonadjacent}) = 1 - P(\text{couple adjacent})

P(\text{couple nonadjacent}) = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}

> [!FORMULA]
> The complementary probability rule states that for any event $E$:
> $$ P(E') = 1 - P(E) $$
> where $E'$ is the event that $E$ does not occur.

✓Final answer

\boxed{\frac{2}{3}}

←Previous questionQ1 — If the letters of the word ALGORITHM are arranged at random in a row, …→Next questionQ3 — Suppose an integer from 1 through 1000 is chosen at random, find the p…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days