Let nCr - 1 = 28, nCr = 56 and nCr + 1 = 70. Let A(4cost, 4sint), B(2sint, -2cost) and C(3r - n, r2 - n

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q22 jee_main_2024_29_jan_morning Dictionary Rank of a Word

All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is

Numerical Answer. Answer: 553 to 553

Solution

Related Formula

Permutations of n objects with p identical elements = (n!)/(p!)

Core Logic

Alphabetize the letters in the word "GTWENTY": E, G, N, T, T, W, Y. We calculate the number of permutations that alphabetically precede "GTWENTY" by exhaustively scanning dictionary prefixes.

Step 1: Calculate Block Combinations

Words starting with 'E':

Remaining letters {G, N, T, T, W, Y}. We have 6 letters with 'T' repeating twice. Permutations = $(6!)/(2!) = (720)/(2) = 360$ $\frac{6!}{2!} = \frac{720}{2} = 360$.

Words starting with 'G': This locks the first letter. Next alphabetical letter is 'E'.
Starting with 'GE':

Remaining {N, T, T, W, Y}. 5 letters, 'T' repeating. Permutations = $(5!)/(2!) = (120)/(2) = 60$ $\frac{5!}{2!} = \frac{120}{2} = 60$.

Starting with 'GN':

Remaining {E, T, T, W, Y}. 5 letters, 'T' repeating. Permutations = $(5!)/(2!) = 60$ $\frac{5!}{2!} = 60$.

Starting with 'GT': This locks the second letter as well. We iterate through the remaining sorted pool {E, N, T, W, Y}.

-- Starting with 'GTE': Remaining {N, T, W, Y}. No repetitions. Permutations = $4! = 24$ $4! = 24$.

-- Starting with 'GTN': Remaining {E, T, W, Y}. No repetitions. Permutations = $4! = 24$ $4! = 24$.

-- Starting with 'GTT': Remaining {E, N, W, Y}. No repetitions. Permutations = $4! = 24$ $4! = 24$.

-- Starting with 'GTW': This locks the third letter. Iterate through {E, N, T, Y}.

Step 2: Trace Remaining Exact String

We are now tracking the prefix 'GTW'. The remaining letters alphabetically are E, N, T, Y. The target word is precisely built out of these letters in exact alphabetical order: E, then N, then T, then Y. This means "GTWENTY" is the very first word in the 'GTW' block. So, it adds exactly 1 to the count.

Step 3: Sum the Permutations

Total serial number = $360 + 60 + 60 + 24 + 24 + 24 + 1 = 553$ .

Pattern Recognition

When calculating dictionary rank with repeating letters, remember to divide by $p!$ $p!$ only when the repeating letter is roaming freely in the available blanks. If a repeating letter is 'locked' as the current prefix, it no longer acts as a repeater for the remaining slots.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Q26 jee_main_2024_30_january_evening Selection of Objects

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is

Numerical Answer. Answer: 11376 to 11376

Solution

Related Formula

Combinations: ⁿCᵣ = (n!)/(r!(n-r)!)

Core Logic

Total Questions = 20 (A: 8, B: 6, C: 6). Total to attempt = 15. Minimum required from each section = 4.

Base attempt gives: $4 (from A) + 4 (from B) + 4 (from C) = 12$ questions. We have to distribute the remaining $15 - 12 = 3$ $15 - 12 = 3$ questions across the sections A, B, and C. Let the additional questions picked be $x, y, z$ $x, y, z$ for sections A, B, C respectively. Then $x + y + z = 3$ $x + y + z = 3$, with constraints based on the maximum questions per section: A max extra = $8 - 4 = 4 \Rightarrow x \leq 4$ $8 - 4 = 4 \Rightarrow x \le 4$ B max extra = $6 - 4 = 2 \Rightarrow y \leq 2$ $6 - 4 = 2 \Rightarrow y \le 2$ C max extra = $6 - 4 = 2 \Rightarrow z \leq 2$ $6 - 4 = 2 \Rightarrow z \le 2$

Step 1: Identifying Valid Selection Cases

The possible sets of $(x, y, z)$ $(x, y, z)$ are: Case 1: $(1, 1, 1)$ $(1, 1, 1)$ $\Rightarrow$ $\Rightarrow$ Total picks: A(5), B(5), C(5) Case 2: $(2, 1, 0)$ $(2, 1, 0)$ and its permutations (respecting constraints). Valid permutations:

A gets 2, B gets 1, C gets 0 $\Rightarrow$ $\Rightarrow$ A(6), B(5), C(4)
A gets 2, C gets 1, B gets 0 $\Rightarrow$ $\Rightarrow$ A(6), B(4), C(5)
B gets 2, A gets 1, C gets 0 $\Rightarrow$ $\Rightarrow$ A(5), B(6), C(4)
C gets 2, A gets 1, B gets 0 $\Rightarrow$ $\Rightarrow$ A(5), B(4), C(6)

(Note: B(2), C(1) or C(2), B(1) are not allowed if it forces A to take 0, wait, A gets 0 means A(4). A is allowed to have 4.) Let's check permutations of $(2, 1, 0)$ $(2, 1, 0)$:

A(4), B(6), C(5) [x=0, y=2, z=1]
A(4), B(5), C(6) [x=0, y=1, z=2]

Case 3: $(3, 0, 0)$ $(3, 0, 0)$ and permutations. Since $y \leq 2$ $y \le 2$ and $z \leq 2$ $z \le 2$, only $x$ $x$ can be 3. So, $x=3, y=0, z=0$ $x=3, y=0, z=0$ $\Rightarrow$ $\Rightarrow$ A(7), B(4), C(4).

Step 2: Calculating Combinations per Case

Let's list all valid final section breakdowns $(A, B, C)$ $(A, B, C)$:

$(5, 5, 5) \Rightarrow ⁸C₅ \cdot ⁶C₅ \cdot ⁶C₅ = 56 \cdot 6 \cdot 6 = 2016$
$(6, 5, 4) \Rightarrow ⁸C₆ \cdot ⁶C₅ \cdot ⁶C₄ = 28 \cdot 6 \cdot 15 = 2520$
$(6, 4, 5) \Rightarrow ⁸C₆ \cdot ⁶C₄ \cdot ⁶C₅ = 28 \cdot 15 \cdot 6 = 2520$
$(5, 6, 4) \Rightarrow ⁸C₅ \cdot ⁶C₆ \cdot ⁶C₄ = 56 \cdot 1 \cdot 15 = 840$
$(5, 4, 6) \Rightarrow ⁸C₅ \cdot ⁶C₄ \cdot ⁶C₆ = 56 \cdot 15 \cdot 1 = 840$
$(4, 6, 5) \Rightarrow ⁸C₄ \cdot ⁶C₆ \cdot ⁶C₅ = 70 \cdot 1 \cdot 6 = 420$
$(4, 5, 6) \Rightarrow ⁸C₄ \cdot ⁶C₅ \cdot ⁶C₆ = 70 \cdot 6 \cdot 1 = 420$
$(7, 4, 4) \Rightarrow ⁸C₇ \cdot ⁶C₄ \cdot ⁶C₄ = 8 \cdot 15 \cdot 15 = 1800$

Step 3: Summing the Total Ways

Total ways = $2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$ Total ways = $2016 + 5040 + 1680 + 840 + 1800 = 11376$

Pattern Recognition

Combinatorial distribution with rigid lower bounds is solved by shifting the baseline. Allocate the minimums immediately ( $4+4+4=12$ $4+4+4=12$), then distribute the remaining items via casework ensuring upper capacities aren't breached.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q1 jee_main_2024_31_jan_evening Distribution of Identical Objects

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

A. $406$ $406$
B. $130$ $130$
C. $142$ $142$
D. $136$ $136$

Solution

Related Formula

Ways to distribute n identical objects among r persons = n+r-1Cᵣ₋₁

Core Logic

First, distribute 2 apples to each of the 3 children to satisfy the minimum requirement. Remaining apples = $21 - 3 \times 2 = 15$ $21 - 3 \times 2 = 15$. Now distribute the remaining 15 identical apples among the 3 children without restrictions.

Number of ways = ¹⁵⁺³⁻¹C₃₋₁ = ¹⁷C₂

= (17 \times 16)/(2) = 136

Pattern Recognition

Beggar's Method: For $x₁+x₂+ +xᵣ = n$ $x_1+x_2+\dots+x_r = n$ with $xᵢ \geq k$ $x_i \ge k$, pre-allocate $k$ $k$ to each and apply standard distribution formula on remainder.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q17 jee_main_2024_31_jan_evening Combinations and Permutations Formula

If for some

m, n

;

⁶Cm + 2(⁶Cm+1) + ⁶Cm+2 > ⁸C₃

and

ⁿ⁻¹P₃ : ⁿP₄ = 1:8

, then

ⁿPm+1 + ⁿ⁺¹Cm

is equal to

A. $380$ $380$
B. $376$ $376$
C. $384$ $384$
D. $372$ $372$

Solution

Related Formula

ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Core Logic

Simplify the binomial combination:

⁶Cm + 2(⁶Cm+1) + ⁶Cm+2 = (⁶Cm + ⁶Cm+1) + (⁶Cm+1 + ⁶Cm+2)

Using Pascal's rule, this becomes:

⁷Cm+1 + ⁷Cm+2 = ⁸Cm+2

Given condition: $⁸Cm+2 > ⁸C₃ = 56$ $^8C_{m+2} > ^8C_3 = 56$. For $N=8$ $N=8$, the central combinations yield the maximum value: $⁸C₄ = 70$ $^8C_4 = 70$. Others like $⁸C₅ = 56$ $^8C_5 = 56$, which is not strictly greater than 56. So $m + 2 = 4 m = 2$ $m + 2 = 4 \implies m = 2$.

Solve the permutations ratio:

ⁿ⁻¹P₃ⁿP₄ = (1)/(8)

((n-1)(n-2)(n-3))/(n(n-1)(n-2)(n-3)) = (1)/(8) (1)/(n) = (1)/(8) n = 8

Calculate the target expression:

ⁿPm+1 + ⁿ⁺¹Cm = ⁸P₃ + ⁹C₂

= (8 \times 7 \times 6) + (9 \times 8)/(2)

= 336 + 36 = 372

Pattern Recognition

Binomial coefficient reduction using Pascal's triangle quickly collapses expanded $nCr$ $nCr$ sums.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Q23 jee_main_2024_31_jan_morning Word Formation

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to

Numerical Answer. Answer: 3734 to 3734

Solution

Core Logic

Letters in 'DISTRIBUTION': I(3), T(2), D, S, R, B, U, O, N. There are 9 distinct letters.

Step 1: Case Analysis

Case 1: 3 alike, 1 distinct Selection: Choose the letter 'I' ( $¹C₁$ $^1C_1$) and 1 from the remaining 8 distinct letters ( $⁸C₁$ $^8C_1$). Arrangement: $⁸C₁ \times (4!)/(3!) = 8 \times 4 = 32$ .

Case 2: 2 alike of one kind, 2 alike of another kind Since only 'I' and 'T' appear at least twice, we must choose both. Arrangement: $²C₂ \times (4!)/(2!2!) = 1 \times 6 = 6$ .

Step 2: Further Cases

Case 3: 2 alike, 2 distinct Selection: Choose 1 from the 2 repeated sets ( $²C₁$ $^2C_1$) and 2 from the remaining 8 distinct letters ( $⁸C₂$ $^8C_2$). Arrangement: $²C₁ \times ⁸C₂ \times (4!)/(2!) = 2 \times 28 \times 12 = 672$ .

Case 4: All 4 distinct Selection: Choose 4 from the 9 distinct letters ( $⁹C₄$ $^9C_4$). Arrangement: $⁹C₄ \times 4! = 126 \times 24 = 3024$ .

Step 3: Total Words

Total = $3024 + 672 + 6 + 32 = 3734$ $3024 + 672 + 6 + 32 = 3734$.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Solution & Explanation

Related Formula

Core Logic

Step 1: Locating Vertices and Centroid Locus

Step 2: Squaring and Summing Trig Components

Pattern Recognition

Chapter Mix

More Permutations and Combinations Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Block Combinations

Step 2: Trace Remaining Exact String

Step 3: Sum the Permutations

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identifying Valid Selection Cases

Step 2: Calculating Combinations per Case

Step 3: Summing the Total Ways

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Case Analysis

Step 2: Further Cases

Step 3: Total Words

Chapter Mix

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