Solution & Explanation
### Related Formula
textPermutations of n text objects with p text identical elements = fracn!p!$$\text{Permutations of } n \text{ objects with } p \text{ identical elements } = \frac{n!}{p!}$$
### 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
1. Words starting with 'E':
Remaining letters {G, N, T, T, W, Y}. We have 6 letters with 'T' repeating twice.
Permutations = frac6!2! = frac7202 = 360$\frac{6!}{2!} = \frac{720}{2} = 360$.
2. 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 = frac5!2! = frac1202 = 60$\frac{5!}{2!} = \frac{120}{2} = 60$.
- Starting with 'GN':
Remaining {E, T, T, W, Y}. 5 letters, 'T' repeating.
Permutations = frac5!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$360 + 60 + 60 + 24 + 24 + 24 + 1 = 553$.
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
More Permutations and Combinations Previous-Year Questions — Page 2
Q58
jee_main_2025_03_april_evening
Geometry Problems
Line L_1$L_1$ of slope 2$2$ and line L_2$L_2$ of slope frac12$\frac{1}{2}$ intersect at the origin O$O$. In the first quadrant, P_1, P_2, dots, P_12$P_1, P_2, \dots, P_{12}$ are 12$12$ points on line L_1$L_1$ and Q_1, Q_2, dots, Q_9$Q_1, Q_2, \dots, Q_9$ are 9$9$ points on line L_2$L_2$. Then the total number of triangles, that can be formed having vertices at three of the 22$22$ points O, P_1, P_2, dots, P_12, Q_1, Q_2, dots, Q_9$O, P_1, P_2, \dots, P_{12}, Q_1, Q_2, \dots, Q_9$, is:
- A. 1080$1080$
- B. 1134$1134$
- C. 1026$1026$
- D. 1188$1188$
Solution
### Related Formula
Number of ways to choose 3$3$ points out of N$N$ points is ^N C_3$^N C_3$.
If any points are collinear, choosing 3$3$ points from those collinear points will form a straight line instead of a triangle.
### Core Logic
Total number of points = 1$1$ (origin O$O$) + 12$12$ (on L_1$L_1$) + 9$9$ (on L_2$L_2$) = 22$22$ points.
Triangles are formed by choosing any 3$3$ points except those that are collinear.
### Step 1: Identifying Collinear Sets
1. The set of points lying on L_1$L_1$ includes O, P_1, dots, P_12$O, P_1, \dots, P_{12}$, which is 13$13$ collinear points.
Number of collinear combinations = ^13 C_3$^{13} C_3$
2. The set of points lying on L_2$L_2$ includes O, Q_1, dots, Q_9$O, Q_1, \dots, Q_9$, which is 10$10$ collinear points.
Number of collinear combinations = ^10 C_3$^{10} C_3$
### Step 2: Triangle Calculation
Total Triangles = Total choices of 3 points - Collinear choices on L_1$L_1$ - Collinear choices on L_2$L_2$
textTotal = ^22 C_3 - ^13 C_3 - ^10 C_3$$\text{Total } = ^{22} C_3 - ^{13} C_3 - ^{10} C_3$$
Calculating individual combinations:
- ^22 C_3 = frac22 times 21 times 203 times 2 times 1 = 1540$^{22} C_3 = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 1540$
- ^13 C_3 = frac13 times 12 times 113 times 2 times 1 = 286$^{13} C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$
- ^10 C_3 = frac10 times 9 times 83 times 2 times 1 = 120$^{10} C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
textTriangles = 1540 - 286 - 120 = 1134$$\text{Triangles} = 1540 - 286 - 120 = 1134$$
### 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)$$\text{Triangles} = (^{12} C_2 \times ^9 C_1) + (^9 C_2 \times ^{12} C_1) + (1 \times ^{12} C_1 \times ^9 C_1)$$
= (66 times 9) + (36 times 12) + (108) = 594 + 432 + 108 = 1134$$= (66 \times 9) + (36 \times 12) + (108) = 594 + 432 + 108 = 1134$$
### 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$r$ items from a pool of n$n$ distinct objects:
^nC_r = fracn!r!(n-r)!$$^nC_r = \frac{n!}{r!(n-r)!}$$
### Core Logic
Total required players = 10$10$.
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$10 - 2 = 8$ players.
- Available remaining pool:
- Batsmen available = 7 - 1 = 6$7 - 1 = 6$ batsmen.
- Bowlers available = 6 - 1 = 5$6 - 1 = 5$ bowlers.
Adjusted structural constraints for the remaining 8 slots:
- Needs at least 4 - 1 = 3$4 - 1 = 3$ more batsmen.
- Needs at least 4 - 1 = 3$4 - 1 = 3$ more bowlers.
### Step 1: Set Up Case Combinations
Let x$x$ be the number of additional batsmen and y$y$ be the number of additional bowlers selected, where x + y = 8$x + y = 8$ with x ge 3$x \ge 3$ and y ge 3$y \ge 3$.
Possible case matches:
- **Case 1**: 5$5$ batsmen and 3$3$ bowlers (x=5, y=3$x=5, y=3$)
- **Case 2**: 4$4$ batsmen and 4$4$ bowlers (x=4, y=4$x=4, y=4$)
- **Case 3**: 3$3$ batsmen and 5$5$ bowlers (x=3, y=5$x=3, y=5$)
### Step 2: Calculate Each Case Value
1. For Case 1:
textWays = ^6C_5 times ^5C_3 = 6 times 10 = 60$$\text{Ways} = ^6C_5 \times ^5C_3 = 6 \times 10 = 60$$
2. For Case 2:
textWays = ^6C_4 times ^5C_4 = 15 times 5 = 75$$\text{Ways} = ^6C_4 \times ^5C_4 = 15 \times 5 = 75$$
3. For Case 3:
textWays = ^6C_3 times ^5C_5 = 20 times 1 = 20$$\text{Ways} = ^6C_3 \times ^5C_5 = 20 \times 1 = 20$$
### Step 3: Compute Total Ways
Sum the combinations across all valid exhaustive paths:
textTotal Ways = 60 + 75 + 20 = 155$$\text{Total Ways} = 60 + 75 + 20 = 155$$
### 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$n$) and the final destination requirement count (r$r$) 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$n \geq 2$ , 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
