There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturer is to be forme

NCERT Exemplar · Q62

10

(a) In how many ways committee can be formed;

(b) In how many ways a particular professor is included;

(d) In how many ways a particular lecturer is excluded.

C_2

(i)

{}^{10}C_{2} \times {}^{19}C_{3}

;

(ii)

{}^{10}C_{2} \times {}^{19}C_{2}

;

(iii)

{}^{9}C_{1} \times {}^{20}C_{3}

;

(iv)

{}^{10}C_{2} \times {}^{20}C_{3}

Yanam BieapShort· 2mImportance★★★★★est

98% · 128/130 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 — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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

This problem involves forming a committee with specific conditions, requiring the application of combinations. The key is to adjust the total number of available people and the number of people to be chosen based on whether a particular individual is included or excluded. The final matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

When forming a committee, the order in which members are selected does not matter. This means we are dealing with combinations, not permutations. A combination is a selection of items from a larger set where the order of selection is irrelevant.

k

The problem asks us to form a committee of $2$ professors and $3$ lecturers. We have $10$ professors and $20$ lecturers in total. For each scenario, we will determine the number of ways to choose professors and lecturers separately, then multiply these numbers together using the fundamental principle of counting, as the choices for professors and lecturers are independent events.

Let's break down each item in Column $C_1$ :

Item (a): In how many ways a committee can be formed?

This is the most straightforward case, where no specific conditions are imposed on individual members.
- Choosing Professors: We need to select $2$ professors from the available $10$ professors. The number of ways to do this is ${}^{10}C_2$ .
- Choosing Lecturers: We need to select $3$ lecturers from the available $20$ lecturers. The number of ways to do this is ${}^{20}C_3$ .
- Total Ways: Since the selection of professors and lecturers are independent, we multiply the number of ways for each. Total ways = ${}^{10}C_2 \times {}^{20}C_3$ .
- This matches option (iv) in Column $C_2$ .
Item (b): In how many ways a particular professor is included?

Here, one specific professor must be part of the committee.
- Choosing Professors: Since one particular professor is already included, we effectively have $1$ professor already selected. We still need to select $2 - 1 = 1$ more professor. The pool of available professors for this remaining selection is now $10 - 1 = 9$ (the total professors minus the one already selected). So, we choose $1$ professor from $9$ . The number of ways is ${}^{9}C_1$ .
- Choosing Lecturers: There are no special conditions for lecturers, so we select $3$ lecturers from the available $20$ lecturers. The number of ways is ${}^{20}C_3$ .
- Total Ways: Multiply the ways for professors and lecturers. Total ways = ${}^{9}C_1 \times {}^{20}C_3$ .
- This matches option (iii) in Column $C_2$ .
Item (c): In how many ways a particular lecturer is included?

Similar to item (b), one specific lecturer must be part of the committee.
- Choosing Professors: There are no special conditions for professors, so we select $2$ professors from the available $10$ professors. The number of ways is ${}^{10}C_2$ . …

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

←Previous questionQ61 — Five boys and five girls form a line. Find the number of ways of makin…→Next questionQ63 — Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits i…

Q.There are 101010 professors and 202020 lecturers out of whom a committee of 222 professors and 333 lecturer is to be formed. Match each item in Column C1C_1C1​ with its correct answer in Column C2C_2C2​. C1C_1C1​:

🎓Unlock everything free for 14 days

Q.There are $10$ professors and $20$ lecturers out of whom a committee of $2$ professors and $3$ lecturer is to be formed. Match each item in Column $C_1$ with its correct answer in Column $C_2$ . $C_1$ :

Unlock everything free for 14 days