Solution & Explanation
### Related Formula
textCombinations: ^nC_r = fracn!r!(n-r)!$\text{Combinations: } {}^nC_r = \frac{n!}{r!(n-r)!}$
### 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 (textfrom A) + 4 (textfrom B) + 4 (textfrom C) = 12$4 (\text{from A}) + 4 (\text{from B}) + 4 (\text{from C}) = 12$ 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 le 4$8 - 4 = 4 \Rightarrow x \le 4$
B max extra = 6 - 4 = 2 Rightarrow y le 2$6 - 4 = 2 \Rightarrow y \le 2$
C max extra = 6 - 4 = 2 Rightarrow z le 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 le 2$y \le 2$ and z le 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)$(A, B, C)$:
1) (5, 5, 5) Rightarrow ^8C_5 cdot ^6C_5 cdot ^6C_5 = 56 cdot 6 cdot 6 = 2016$(5, 5, 5) \Rightarrow {}^8C_5 \cdot {}^6C_5 \cdot {}^6C_5 = 56 \cdot 6 \cdot 6 = 2016$
2) (6, 5, 4) Rightarrow ^8C_6 cdot ^6C_5 cdot ^6C_4 = 28 cdot 6 cdot 15 = 2520$(6, 5, 4) \Rightarrow {}^8C_6 \cdot {}^6C_5 \cdot {}^6C_4 = 28 \cdot 6 \cdot 15 = 2520$
3) (6, 4, 5) Rightarrow ^8C_6 cdot ^6C_4 cdot ^6C_5 = 28 cdot 15 cdot 6 = 2520$(6, 4, 5) \Rightarrow {}^8C_6 \cdot {}^6C_4 \cdot {}^6C_5 = 28 \cdot 15 \cdot 6 = 2520$
4) (5, 6, 4) Rightarrow ^8C_5 cdot ^6C_6 cdot ^6C_4 = 56 cdot 1 cdot 15 = 840$(5, 6, 4) \Rightarrow {}^8C_5 \cdot {}^6C_6 \cdot {}^6C_4 = 56 \cdot 1 \cdot 15 = 840$
5) (5, 4, 6) Rightarrow ^8C_5 cdot ^6C_4 cdot ^6C_6 = 56 cdot 15 cdot 1 = 840$(5, 4, 6) \Rightarrow {}^8C_5 \cdot {}^6C_4 \cdot {}^6C_6 = 56 \cdot 15 \cdot 1 = 840$
6) (4, 6, 5) Rightarrow ^8C_4 cdot ^6C_6 cdot ^6C_5 = 70 cdot 1 cdot 6 = 420$(4, 6, 5) \Rightarrow {}^8C_4 \cdot {}^6C_6 \cdot {}^6C_5 = 70 \cdot 1 \cdot 6 = 420$
7) (4, 5, 6) Rightarrow ^8C_4 cdot ^6C_5 cdot ^6C_6 = 70 cdot 6 cdot 1 = 420$(4, 5, 6) \Rightarrow {}^8C_4 \cdot {}^6C_5 \cdot {}^6C_6 = 70 \cdot 6 \cdot 1 = 420$
8) (7, 4, 4) Rightarrow ^8C_7 cdot ^6C_4 cdot ^6C_4 = 8 cdot 15 cdot 15 = 1800$(7, 4, 4) \Rightarrow {}^8C_7 \cdot {}^6C_4 \cdot {}^6C_4 = 8 \cdot 15 \cdot 15 = 1800$
### Step 3: Summing the Total Ways
Total ways = 2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$
Total ways = 2016 + 5040 + 1680 + 840 + 1800 = 11376$2016 + 5040 + 1680 + 840 + 1800 = 11376$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Permutations and Combinations
More Permutations and Combinations Previous-Year Questions — Page 2
Q75
jee_main_2025_07_april_morning
Subsets without Consecutive Elements
For n geq 2$n \geq 2$ , let S_n$S_{n}$ denote the set of all subsets of \1, 2, dots, n\$\{1, 2, \dots, n\}$ with no two consecutive numbers. For example \1, 3, 5\ in mathbfS_6$\{1, 3, 5\} \in \mathbf{S}_6$ , but \1, 2, 4\ notin mathbfS_6$\{1, 2, 4\} \notin \mathbf{S}_6$ . Then n(mathbfS_5)$n(\mathbf{S}_5)$ is equal to
Numerical Answer. Answer: 13 to 13
Solution
### Related Formula
The number of ways to choose r$r$ non-consecutive elements from a set of n$n$ items is given by the binomial formula:
textWays = ^n - r + 1C_r$\text{Ways} = {}^{n - r + 1}C_r$
### Core Logic
We need to find the total number of subsets for n = 5$n = 5$ items without consecutive values.
We break down the options based on the size of the subset (r$r$), ranging from an empty set (r=0$r=0$) up to the maximum possible non-consecutive size (r=3$r=3$).
### Step 1: Compute Combinations for Each Size
1. Subsets containing **no elements** (r = 0$r = 0$):
textWays = 1 quad (textThe empty set emptyset)$\text{Ways} = 1 \quad (\text{The empty set } \emptyset)$
2. Subsets containing **exactly 1 element** (r = 1$r = 1$):
textWays = ^5 - 1 + 1C_1 = ^5C_1 = 5 quad (\1\, \2\, \3\, \4\, \5\)$\text{Ways} = {}^{5 - 1 + 1}C_1 = {}^5C_1 = 5 \quad (\{1\}, \{2\}, \{3\}, \{4\}, \{5\})$
3. Subsets containing **exactly 2 elements** (r = 2$r = 2$):
textWays = ^5 - 2 + 1C_2 = ^4C_2 = 6$\text{Ways} = {}^{5 - 2 + 1}C_2 = {}^4C_2 = 6$
4. Subsets containing **exactly 3 elements** (r = 3$r = 3$):
textWays = ^5 - 3 + 1C_3 = ^3C_3 = 1 quad (\1, 3, 5\)$\text{Ways} = {}^{5 - 3 + 1}C_3 = {}^3C_3 = 1 \quad (\{1, 3, 5\})$
### Step 2: Sum the Total Counts
Add all possible valid non-consecutive subset paths together:
textTotal Subsets n(S_5) = 1 + 5 + 6 + 1 = 13$\text{Total Subsets } n(S_5) = 1 + 5 + 6 + 1 = 13$
### Pattern Recognition
Shortcut: The total number of non-consecutive subsets for a set of size n$n$ follows the Fibonacci sequence pattern: F_n+2$F_{n+2}$. For n=5$n=5$, the value is the 7th Fibonacci number, which is exactly 13$13$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q63
jee_main_2025_08_april_evening
Geometry Problems in Combinatorics
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is
- A. 230$230$
- B. 220$220$
- C. 200$200$
- D. 210$210$
Solution
### Related Formula
textNet Triangles = binomn3 - binomm3$\text{Net Triangles} = \binom{n}{3} - \binom{m}{3}$
### Core Logic
To calculate valid un-collapsed geometric triangles, find total ways to select 3 distinct coordinate indicators from the set and remove choices restricted entirely within the inline row sequence.
### Step 1: Compute Full Dynamic Combinations
Choosing 3 general elements from the array size of 12:
binom123 = frac12 times 11 times 103 times 2 times 1 = 220$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$
### Step 2: Isolate Internal Flattened Collinear Triplets
Choosing 3 items completely bundled inside the collinear group row of size 5:
binom53 = 10$\binom{5}{3} = 10$
### Step 3: Deduce Triangles
textTriangles Generated = 220 - 10 = 210$\text{Triangles Generated} = 220 - 10 = 210$
### Pattern Recognition
Points in a straight line cannot create spatial enclosing fields. Subtracting localized combinations from generalized permutations accounts for structural constraints.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q60
jee_main_2025_29_jan_evening
Rank of a Word in Dictionary
If all the words with or without meaning made using all the letters of the word “KANPUR” are arranged as in a dictionary, then the word at 440^textth$440^{\text{th}}$ position in this arrangement, is :
- A. textPRNAKU$\text{PRNAKU}$
- B. textPRKANU$\text{PRKANU}$
- C. textPRKAUN$\text{PRKAUN}$
- D. textPRNAUK$\text{PRNAUK}$
Solution
### Related Formula
Number of permutations of n$n$ distinct letters is given by:
textPermutations = n!$\text{Permutations} = n!$
### Core Logic
Sort the distinct letters of "KANPUR" alphabetically:
textA, K, N, P, R, U$\text{A, K, N, P, R, U}$
Track alphabetical structural sets row by row:
Words starting with A: 5! = 120$5! = 120$
Words starting with K: 5! = 120$5! = 120$ (Cumulative: 240)
Words starting with N: 5! = 120$5! = 120$ (Cumulative: 360)
### Step 1: Parse the Next Character Layer
We need the 440th word, so the next block begins with P:
Words starting with PA: 4! = 24$4! = 24$ (Cumulative: 384)
Words starting with PK: 4! = 24$4! = 24$ (Cumulative: 408)
Words starting with PN: 4! = 24$4! = 24$ (Cumulative: 432)
### Step 2: Reach the Targeted Count
Remaining difference to hit 440 is 440 - 432 = 8$440 - 432 = 8$ words.
The next group starts with PR. Alphabetical listings within PR:
Words starting with PRA: 3! = 6$3! = 6$ (Cumulative: 438)
Now list alphabetically within PRK:
439th word: PRKANU
440th word: PRKAUN
### Pattern Recognition
Factorial blocks eliminate large numbers quickly. Once the cumulative total lands within a narrow range of the answer, switch over to writing combinations down manually to ensure high accuracy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q51
jee_main_2025_28_jan_morning
Permutations under Restrictions
The number of different
5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is
(1) 4608
(2) 5720
(3) 5719
(4) 4607
- A. 4608
- B. 5720
- C. 5719
- D. 4607
Solution
### Related Formula
For a 5-digit number, total permutations with repetition allowed for n$n$ digits is given by:
textTotal Cases = d_1 times d_2 times d_3 times d_4 times d_5$\text{Total Cases} = d_1 \times d_2 \times d_3 \times d_4 \times d_5$
### Core Logic
We need 5-digit numbers greater than 50000 using digits \0, 1, 2, 3, 4, 5, 6, 7\$\{0, 1, 2, 3, 4, 5, 6, 7\}$ under the restriction d_1 + d_5 le 8$d_1 + d_5 \le 8$.
Let's analyze the pairs (d_1, d_5)$(d_1, d_5)$ where d_1 in \5, 6, 7\$d_1 \in \{5, 6, 7\}$:
Case I: d_1 = 5 Rightarrow d_5 in \0, 1, 2, 3\$d_1 = 5 \Rightarrow d_5 \in \{0, 1, 2, 3\}$ (4 options)
Case II: d_1 = 6 Rightarrow d_5 in \0, 1, 2\$d_1 = 6 \Rightarrow d_5 \in \{0, 1, 2\}$ (3 options)
Case III: d_1 = 7 Rightarrow d_5 in \0, 1\$d_1 = 7 \Rightarrow d_5 \in \{0, 1\}$ (2 options)
Total choices for the first and last digits combined = 4 + 3 + 2 = 9$4 + 3 + 2 = 9$ pairs.
### Step 1: Calculating Intermediate Choices
The middle three digits (d_2, d_3, d_4$d_2, d_3, d_4$) have no restrictions and can each be chosen from any of the 8 available digits.
textNumber of ways = 9 times (8 times 8 times 8) = 4608$\text{Number of ways} = 9 \times (8 \times 8 \times 8) = 4608$
### Step 2: Subtracting Boundary Conditions
Since the question specifies numbers strictly greater than 50000, we must check if 50000 is included in our count.
For d_1=5$d_1=5$ and d_5=0$d_5=0$, setting d_2=d_3=d_4=0$d_2=d_3=d_4=0$ gives exactly 50000, which is included in the 4608 count.
textTotal numbers = 4608 - 1 = 4607$\text{Total numbers} = 4608 - 1 = 4607$
### Pattern Recognition
Always look carefully at edge constraints like "greater than". Counting the number 50000 explicitly avoids typical off-by-one errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Permutations and Combinations
Q67
jee_main_2025_28_jan_morning
Combinatorial Coefficients and Locus
Let ^nC_r - 1 = 28, ^nC_r = 56${}^nC_{r - 1} = 28, {}^nC_r = 56$ and ^nC_r + 1 = 70${}^nC_{r + 1} = 70$. Let A(4cost, 4sint), B(2sint, -2cost) and C(3r - n, r^2 - n - 1$r^2 - n - 1$) be the vertices of a triangle ABC, where t is a parameter. If (3x - 1)^2 + (3y)^2 = alpha$(3x - 1)^2 + (3y)^2 = \alpha$, is the locus of the centroid of triangle ABC, then alpha$\alpha$ equals:
(1) 20
(2) 8
(3) 6
(4) 18
Solution
### Related Formula
Consecutive combinations ratio property:
frac^nC_r-1^nC_r = fracrn-r+1$\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{r}{n-r+1}$
### Core Logic
Setting up ratios between consecutive given coefficients:
frac2856 = frac12 = fracrn-r+1 implies 3r = n + 1 quad dots (1)$\frac{28}{56} = \frac{1}{2} = \frac{r}{n-r+1} \implies 3r = n + 1 \quad \dots (1)$
frac5670 = frac45 = fracr+1n-r implies 9r = 4n - 5 quad dots (2)$\frac{56}{70} = \frac{4}{5} = \frac{r+1}{n-r} \implies 9r = 4n - 5 \quad \dots (2)$
Solving equations (1) and (2) gives r = 3$r = 3$ and n = 8$n = 8$.
### Step 1: Locating Vertices and Centroid Locus
Substituting values for point C$C$ gives C(1,0)$C(1,0)$.
Let the centroid coordinates be (x,y)$(x,y)$:
3x = 4cos t + 2sin t + 1 implies 3x - 1 = 4cos t + 2sin t$3x = 4\cos t + 2\sin t + 1 \implies 3x - 1 = 4\cos t + 2\sin t$
3y = 4sin t - 2cos t + 0 implies 3y = 4sin t - 2cos t$3y = 4\sin t - 2\cos t + 0 \implies 3y = 4\sin t - 2\cos t$
### Step 2: Squaring and Summing Trig Components
Squaring and adding both parametric tracking components eliminates t$t$:
(3x - 1)^2 + (3y)^2 = (4cos t + 2sin t)^2 + (4sin t - 2cos t)^2 = 16 + 4 = 20$(3x - 1)^2 + (3y)^2 = (4\cos t + 2\sin t)^2 + (4\sin t - 2\cos t)^2 = 16 + 4 = 20$
Thus, alpha = 20$\alpha = 20$.
### Pattern Recognition
Symmetric parameter sets of form (Acos t + Bsin t)^2 + (Asin t - Bcos t)^2$(A\cos t + B\sin t)^2 + (A\sin t - B\cos t)^2$ collapse instantly into A^2 + B^2$A^2 + B^2$ via basic Pythagorean identities.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Class 11 Maths: Permutations and Combinations
