A candidate is required to answer $7$ questions out of $12$ questions, which are divided into two groups, each containing $6$ questions. He is not permitted to attempt more than $5$ questions from either group. Find the number of different ways of doing questions.

> [!TLDR] > The candidate must choose $7$ questions from two groups of $6$ each, with at most $5$ from any group. This forces exactly two valid distributions: $(5, 2)$ or $(2, 5)$. The total number of ways is **$\boxed{780}$**. The heart of this problem lies in recognizing that the constraint "at most $5$ from either group" isn't just a ceiling—it actually determines which distributions are possible. When you need $7$ questions total and can't take more than $5$ from one group, the arithmetic forces your hand. Let's denote the two groups as Group A and Group B, each containing $6$ questions. If the candidate selects $x$ questions from Group A, then he must select $7 - x$ questions from Group B. The constraints are: - $0 \le x \le 6$ (can't select more than available) - $x \le 5$ (at most $5$ from Group A) - $7 - x \le 5$ (at most $5$ from Group B), which gives $x \ge 2$ Combining these: $2 \le x \le 5$. But we also need $7 - x \le 6$ (can't select more than available from Group B), which gives $x \ge 1$. This is already satisfied. So the valid values are $x \in \{2, 3, 4, 5\}$. Now let's check each case: 1. **Case $x = 2$:** Select $2$ from Group A and $5$ from Group B. The number of ways is $\binom{6}{2} \times \binom{6}{5}$. We have $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ and $\binom{6}{5} = 6$. Total: $15 \times 6 = 90$ ways. 2. **Case $x = 3$:** Select $3$ from Group A and $4$ from Group B. The number of ways is $\binom{6}{3} \times \binom{6}{4}$. We have $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ and $\binom{6}{4} = \binom{6}{2} = 15$. Total: $20 \times 15 = 300$ ways. 3. **Case $x = 4$:** Select $4$ from Group A and $3$ from Group B. The number of ways is $\binom{6}{4} \times \binom{6}{3}$. We have $\binom{6}{4} = 15$ and $\binom{6}{3} = 20$. Total: $15 \times 20 = 300$ ways. 4. **Case $x = 5$:** Select $5$ from Group A and $2$ from Group B.

A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each con

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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

Watch out

Do not use the permutation formula when order doesn't matter. For example, choosing 3 friends from a group of 5 to form a committee — here the order of selection is irrelevant. That's a combination, not a permutation. Permutations are for ordered arrangements (like rankings, seating orders, passwords where position matters).

Quick Examples

Scenario	Calculation	Answer
Arranging 4 different trophies on a shelf	$4!$	24
Number of 3-digit codes from digits 1–9 (no digit repeated)	$P(9,3) = 9 \times 8 \times 7$	504
Seating 5 people in 5 chairs	$5!$	120
Assigning gold, silver, bronze medals to 8 runners	$P(8,3) = 8 \times 7 \times 6$	336

The Bottom Line

Permutations without repetition answer the question: "In how many different ordered ways can I arrange a set of distinct items, using each item at most once?" The answer is always a product of decreasing integers, starting from $n$ and going down $r$ steps. When $r = n$ , it's simply $n!$ .

Permutations Without Repetition is introduced in the NCERT Class 11 Mathematics Permutations and Combinations chapter, and it is exactly the kind of topic students look up when searching "permutations formula class 11 maths" or "arrangement of distinct objects important questions". It also forms the basis for many JEE Main and state CET counting problems that ask you to arrange distinct items without repeating any of them.

Concept: Constrained selection with complementary counting

The candidate must choose 7 questions from 12 (two groups of 6 each), but cannot take more than 5 from any single group. This constraint forces a specific distribution.

If he takes $k$ questions from Group A, he takes $7-k$ from Group B. The constraint "at most 5 from either group" means:

$k \leq 5$ and $7-k \leq 5$ , which gives $2 \leq k \leq 5$

The valid distributions are $(2,5)$ , $(3,4)$ , $(4,3)$ , and $(5,2)$ questions from Groups A and B respectively.

For each distribution, the number of ways is $\binom{6}{k} \times \binom{6}{7-k}$ :

Distribution	Ways
$(2,5)$	$\binom{6}{2} \times \binom{6}{5} = 15 \times 6 = 90$
$(3,4)$	$\binom{6}{3} \times \binom{6}{4} = 20 \times 15 = 300$
$(4,3)$	$\binom{6}{4} \times \binom{6}{3} = 15 \times 20 = 300$
$(5,2)$	$\binom{6}{5} \times \binom{6}{2} = 6 \times 15 = 90$

Total: $90 + 300 + 300 + 90 = 780$

✓Final answer

\boxed{780}

The candidate must choose $7$ questions from two groups of $6$ each, with at most $5$ from any group. This forces exactly two valid distributions: $(5, 2)$ or $(2, 5)$ . The total number of ways is $\boxed{780}$ .

The heart of this problem lies in recognizing that the constraint "at most $5$ from either group" isn't just a ceiling—it actually determines which distributions are possible. When you need $7$ questions total and can't take more than $5$ from one group, the arithmetic forces your hand.

Let's denote the two groups as Group A and Group B, each containing $6$ questions. If the candidate selects $x$ questions from Group A, then he must select $7 - x$ questions from Group B.

The constraints are:

$0 \le x \le 6$ (can't select more than available)
$x \le 5$ (at most $5$ from Group A)
$7 - x \le 5$ (at most $5$ from Group B), which gives $x \ge 2$

Combining these: $2 \le x \le 5$ .

But we also need $7 - x \le 6$ (can't select more than available from Group B), which gives $x \ge 1$ . This is already satisfied.

So the valid values are $x \in \{2, 3, 4, 5\}$ .

Now let's check each case:

Case $x = 2$ : Select $2$ from Group A and $5$ from Group B.

The number of ways is $\binom{6}{2} \times \binom{6}{5}$ .

We have $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ and $\binom{6}{5} = 6$ .

Total: $15 \times 6 = 90$ ways.
Case $x = 3$ : Select $3$ from Group A and $4$ from Group B.

The number of ways is $\binom{6}{3} \times \binom{6}{4}$ .

We have $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ and $\binom{6}{4} = \binom{6}{2} = 15$ .

Total: $20 \times 15 = 300$ ways.
Case $x = 4$ : Select $4$ from Group A and $3$ from Group B.

The number of ways is $\binom{6}{4} \times \binom{6}{3}$ .

We have $\binom{6}{4} = 15$ and $\binom{6}{3} = 20$ .

Total: $15 \times 20 = 300$ ways.
Case $x = 5$ : Select $5$ from Group A and $2$ from Group B.

The number of ways is $\binom{6}{5} \times \binom{6}{2}$ .

We have $\binom{6}{5} = 6$ and $\binom{6}{2} = 15$ .

Total: $6 \times 15 = 90$ ways.

Tip

Notice the symmetry: cases $(2, 5)$ and $(5, 2)$ give the same count, as do $(3, 4)$ and $(4, 3)$ . This symmetry arises because the two groups are identical in size.

Adding all cases together:

90 + 300 + 300 + 90 = 780

✓Final answer

\boxed{780}

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Q.A candidate is required to answer 777 questions out of 121212 questions, which are divided into two groups, each containing 666 questions. He is not permitted to attempt more than 555 questions from either group. Find the number of different ways of doing questions.

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Q.A candidate is required to answer $7$ questions out of $12$ questions, which are divided into two groups, each containing $6$ questions. He is not permitted to attempt more than $5$ questions from either group. Find the number of different ways of doing questions.

Unlock everything free for 14 days