The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty. The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E): Let the rows be R1, R2, and R3, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at most 3 boxes, which cannot fit 5 letters). Thus, we only subtract cases where exactly one row is empty: - Case 1: Row R1 (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R2 and R3: - Case 2: Row R3 (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R1 and R2: - Case 3: Row R2 (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R1 and R3: Subtracting all empty-row cases from the total arrangements: Inclusion-Exclusion Principle: For distribution problems with simple boundary exclusions, subtracting the complement set (invalid configurations) is mathematically much cleaner than calculating all possible partitions of row assignments. Class 11 Mathematics: Permutations and Combinations

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty. The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E): Let the rows be R1, R2, and R3, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty. The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E): Let the rows be R1, R2, and R3, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at most 3 boxes, which cannot fit 5 letters). Thus, we only subtract cases where exactly one row is empty: - Case 1: Row R1 (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R2 and R3: - Case 2: Row R3 (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R1 and R2: - Case 3: Row R2 (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R1 and R3: Subtracting all empty-row cases from the total arrangements: Inclusion-Exclusion Principle: For distribution problems with simple boundary exclusions, subtracting the complement set (invalid configurations) is mathematically much cleaner than calculating all possible partitions of row assignments. Class 11 Mathematics: Permutations and Combinations

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty. The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E): Let the rows be R1, R2, and R3, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row…

Q58 jee_main_2025_03_april_evening Geometry Problems

Line

L_1

of slope

2

and line

L_2

of slope

frac12

intersect at the origin

O

. In the first quadrant,

P_1, P_2, dots, P_12

are

12

points on line

L_1

and

Q_1, Q_2, dots, Q_9

are

9

points on line

L_2

. Then the total number of triangles, that can be formed having vertices at three of the

22

points

O, P_1, P_2, dots, P_12, Q_1, Q_2, dots, Q_9

, is:

A. $1080$ $1080$
B. $1134$ $1134$
C. $1026$ $1026$
D. $1188$ $1188$

Solution

### Related Formula Number of ways to choose

3

points out of

N

points is

^N C_3

. If any points are collinear, choosing

3

points from those collinear points will form a straight line instead of a triangle. ### Core Logic Total number of points =

1

(origin

O

) +

12

(on

L_1

) +

9

(on

L_2

) =

22

points. Triangles are formed by choosing any

3

points except those that are collinear. ### Step 1: Identifying Collinear Sets 1. The set of points lying on

L_1

includes

O, P_1, dots, P_12

, which is

13

collinear points. Number of collinear combinations =

^13 C_3

2. The set of points lying on

L_2

includes

O, Q_1, dots, Q_9

, which is

10

collinear points. Number of collinear combinations =

^10 C_3

### Step 2: Triangle Calculation Total Triangles = Total choices of 3 points - Collinear choices on

L_1

- Collinear choices on

L_2

textTotal = ^22 C_3 - ^13 C_3 - ^10 C_3

Calculating individual combinations: -

^22 C_3 = frac22 times 21 times 203 times 2 times 1 = 1540

-

^13 C_3 = frac13 times 12 times 113 times 2 times 1 = 286

-

^10 C_3 = frac10 times 9 times 83 times 2 times 1 = 120

textTriangles = 1540 - 286 - 120 = 1134

### Pattern Recognition Alternatively, count using partition combinations to avoid large factorials:

textTriangles = (^12 C_2 times ^9 C_1) + (^9 C_2 times ^12 C_1) + (1 times ^12 C_1 times ^9 C_1)

= (66 times 9) + (36 times 12) + (108) = 594 + 432 + 108 = 1134

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

Q57 jee_main_2025_07_april_morning Practical Problems on Combinations

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

A. $165$ $165$
B. $155$ $155$
C. $145$ $145$
D. $135$ $135$

Solution

### Related Formula Number of ways to select

r

items from a pool of

n

distinct objects:

^nC_r = fracn!r!(n-r)!

### Core Logic Total required players =

10

. Constraints: - Minimum 4 batsmen and minimum 4 bowlers. - 1 specific batsman (captain) and 1 specific bowler (vice-captain) are fixed (already selected). Remaining selection required: - Total players left to choose =

10 - 2 = 8

players. - Available remaining pool: - Batsmen available =

7 - 1 = 6

batsmen. - Bowlers available =

6 - 1 = 5

bowlers. Adjusted structural constraints for the remaining 8 slots: - Needs at least

4 - 1 = 3

more batsmen. - Needs at least

4 - 1 = 3

more bowlers. ### Step 1: Set Up Case Combinations Let

x

be the number of additional batsmen and

y

be the number of additional bowlers selected, where

x + y = 8

with

x ge 3

and

y ge 3

. Possible case matches: - **Case 1**:

5

batsmen and

3

bowlers (

x=5, y=3

) - **Case 2**:

4

batsmen and

4

bowlers (

x=4, y=4

) - **Case 3**:

3

batsmen and

5

bowlers (

x=3, y=5

) ### Step 2: Calculate Each Case Value 1. For Case 1:

textWays = ^6C_5 times ^5C_3 = 6 times 10 = 60

2. For Case 2:

textWays = ^6C_4 times ^5C_4 = 15 times 5 = 75

3. For Case 3:

textWays = ^6C_3 times ^5C_5 = 20 times 1 = 20

### Step 3: Compute Total Ways Sum the combinations across all valid exhaustive paths:

textTotal Ways = 60 + 75 + 20 = 155

### Pattern Recognition When specific roles (like captain/vice-captain) are strictly forced into the selection group, always remove them from both the operational choice pool size (

n

) and the final destination requirement count (

r

) before designing your target case distributions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

Q75 jee_main_2025_07_april_morning Subsets without Consecutive Elements

For

n geq 2

, let

S_n

denote the set of all subsets of

\1, 2, dots, n\

with no two consecutive numbers. For example

\1, 3, 5\ in mathbfS_6

, but

\1, 2, 4\ notin mathbfS_6

. Then

n(mathbfS_5)

is equal to

Numerical Answer. Answer: 13 to 13

Solution

### Related Formula The number of ways to choose

r

non-consecutive elements from a set of

n

items is given by the binomial formula:

textWays = ^n - r + 1C_r

### Core Logic We need to find the total number of subsets for

n = 5

items without consecutive values. We break down the options based on the size of the subset (

r

), ranging from an empty set (

r=0

) up to the maximum possible non-consecutive size (

r=3

). ### Step 1: Compute Combinations for Each Size 1. Subsets containing **no elements** (

r = 0

):

textWays = 1 quad (textThe empty set emptyset)

2. Subsets containing **exactly 1 element** (

r = 1

):

textWays = ^5 - 1 + 1C_1 = ^5C_1 = 5 quad (\1\, \2\, \3\, \4\, \5\)

3. Subsets containing **exactly 2 elements** (

r = 2

):

textWays = ^5 - 2 + 1C_2 = ^4C_2 = 6

4. Subsets containing **exactly 3 elements** (

r = 3

):

textWays = ^5 - 3 + 1C_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

### Pattern Recognition Shortcut: The total number of non-consecutive subsets for a set of size

n

follows the Fibonacci sequence pattern:

F_n+2

. For

n=5

, the value is the 7th Fibonacci number, which is exactly

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

### 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

### Step 2: Isolate Internal Flattened Collinear Triplets Choosing 3 items completely bundled inside the collinear group row of size 5:

binom53 = 10

### Step 3: Deduce Triangles

textTriangles 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

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

distinct letters is given by:

textPermutations = n!

### Core Logic Sort the distinct letters of "KANPUR" alphabetically:

textA, K, N, P, R, U

Track alphabetical structural sets row by row: Words starting with A:

5! = 120

Words starting with K:

5! = 120

(Cumulative: 240) Words starting with N:

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

(Cumulative: 384) Words starting with PK:

4! = 24

(Cumulative: 408) Words starting with PN:

4! = 24

(Cumulative: 432) ### Step 2: Reach the Targeted Count Remaining difference to hit 440 is

440 - 432 = 8

words. The next group starts with PR. Alphabetical listings within PR: Words starting with PRA:

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

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:
The grid diagram shows 8 boxes arranged in three horizontal rows of sizes 3, 2, and 3.

Solution & Explanation

Reference Study Guides

More Permutations and Combinations Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Permutations and Combinations Questions — jee_main_2025_02_april_evening

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The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is: The grid diagram shows 8 boxes arranged in three horizontal rows of sizes 3, 2, and 3.

Solution & Explanation

Reference Study Guides

More Permutations and Combinations Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Permutations and Combinations Questions — jee_main_2025_02_april_evening

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The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:
The grid diagram shows 8 boxes arranged in three horizontal rows of sizes 3, 2, and 3.