Let S = \p_1, p_2, ldots, p_10\$S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of the first ten prime numbers. Let A = S cup P$A = S \cup P$, where P$P$ is the set of all possible products of distinct elements of S$S$. Then the number of all ordered pairs (x, y)$(x, y)$, x in S$x \in S$, y in A$y \in A$ such that x$x$ divides y$y$, is ________.
Numerical Answer Type:
Enter a numerical valueAnswer: 5120+4 marks
Solution & Explanation
### Related Formula
The number of subsets of a set containing n$n$ elements is given by the power set formula:
textCount = 2^n$\text{Count} = 2^n$
### Core Logic
Let's analyze the counting criteria for each choice of divisor x in S$x \in S$. Since S$S$ contains 10 elements, there are 10 choices for the prime number x$x$:
Case 1: Elements belonging to \subset S$S$
For a prime x$x$ to divide an entry y in S$y \in S$, y$y$ must be exactly equal to x$x$ itself (since all elements in S$S$ are distinct primes). This yields exactly 1$1$ choice for each prime x$x$.
### Step 1: Count elements belonging to product set P
For a prime x$x$ to divide an entry y in P$y \in P$, where y$y$ is a product of distinct primes from S$S$, the prime x$x$ must be one of the factors included in that product.
To form such a product, x$x$ must be chosen, and the remaining factors can be selected from any combination of the other 9$9$ primes in S$S$. The number of ways to choose subsets from the remaining 9 primes is given by the power set formula:
textWays = 2^9 = 512$\text{Ways} = 2^9 = 512$
### Step 2: Combine and Evaluate Total Ordered Pairs
Sum the valid outcomes from both subsets for a single prime x$x$:
textTotal choices for a fixed x = 1 + 512 = 513 quad text?$\text{Total choices for a fixed } x = 1 + 512 = 513 \quad \text{?}$
Wait, let's re-verify the definition of set P$P$. P$P$ is the set of all possible products of distinct elements of S$S$. Does P$P$ include products of single elements? If a product has only 1 element, it is just the prime itself, which is already in S$S$.
Let's use the alternative \subset framing: an element y in A$y \in A$ corresponds to a non-empty \subset of S$S$ whose elements are multiplied together. For a fixed prime x in S$x \in S$ to divide y$y$, x$x$ must be included in that \subset. The remaining elements of the \subset can be chosen in any way from the remaining 9 primes, which gives:
textTotal subsets containing x = 2^9 = 512$\text{Total subsets containing } x = 2^9 = 512$
Since there are 10 choices for the prime x$x$, the total number of ordered pairs (x,y)$(x,y)$ is:
textTotal Pairs = 10 cdot 2^9 = 10 cdot 512 = 5120$\text{Total Pairs} = 10 \cdot 2^9 = 10 \cdot 512 = 5120$
### Pattern Recognition
Instead of counting the pairs by analyzing values of y$y$ first, reversing the calculation to count based on the number of choices for the divisor x$x$ simplifies the problem into a straightforward power set calculation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Class 11 Mathematics: Sets
Keywords:#combinatorial power set subsets#prime divisibility counting#JEE Main 2025 Morning Q75#ordered pairs combinations
More Permutations and Combinations Previous-Year Questions
Q662025Arrangements
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.
A. 5880
B. 960
C. 840
D. 5760
Solution
### Related Formula
textNumber of arrangements of r text items in n text boxes = binomnr cdot r!$\text{Number of arrangements of } r \text{ items in } n \text{ boxes} = \binom{n}{r} \cdot r!$
### Core Logic
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.
### Step 1: Compute total unrestricted arrangements
The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E):
textTotal unrestricted arrangements = binom85 cdot 5! = 56 cdot 120 = 6720$\text{Total unrestricted arrangements} = \binom{8}{5} \cdot 5! = 56 \cdot 120 = 6720$
### Step 2: Identify and subtract the invalid empty-row cases
Let the rows be R_1$R_1$, R_2$R_2$, and R_3$R_3$, 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 R_1$R_1$ (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R_2$R_2$ and R_3$R_3$:
textWays = binom55 cdot 5! = 120$\text{Ways} = \binom{5}{5} \cdot 5! = 120$
- Case 2: Row R_3$R_3$ (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R_1$R_1$ and R_2$R_2$:
textWays = binom55 cdot 5! = 120$\text{Ways} = \binom{5}{5} \cdot 5! = 120$
- Case 3: Row R_2$R_2$ (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R_1$R_1$ and R_3$R_3$:
textWays = binom65 cdot 5! = 6 cdot 120 = 720$\text{Ways} = \binom{6}{5} \cdot 5! = 6 \cdot 120 = 720$
### Step 3: Calculate the final valid arrangements
Subtracting all empty-row cases from the total arrangements:
textValid arrangements = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760$\text{Valid arrangements} = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760$
### Pattern Recognition
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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q2025Exponent of Prime in a Factorial
The largest n in mathbbN$n \in \mathbb{N}$ such that 3^n$3^n$ divides 50!$50!$ is:
A.21$21$
B.22$22$
C.20$20$
D.23$23$
Solution
### Related Formula
The exponent of a prime p$p$ in N!$N!$ is given by Legendre's formula:
E_p(N!) = left[fracNpright] + left[fracNp^2right] + left[fracNp^3right] + dots$E_p(N!) = \left[\frac{N}{p}\right] + \left[\frac{N}{p^2}\right] + \left[\frac{N}{p^3}\right] + \dots$
### Core Logic
To find the highest power of 3$3$ that divides 50!$50!$, calculate the sum of the greatest integer functions for successive powers of 3$3$ up to 50$50$.
### Step 1: Computation
Applying the formula for N = 50$N = 50$ and p = 3$p = 3$:
E_3(50!) = left[frac503right] + left[frac509right] + left[frac5027right] + left[frac5081right]$E_3(50!) = \left[\frac{50}{3}\right] + \left[\frac{50}{9}\right] + \left[\frac{50}{27}\right] + \left[\frac{50}{81}\right]$E_3(50!) = 16 + 5 + 1 + 0 = 22$E_3(50!) = 16 + 5 + 1 + 0 = 22$
### Pattern Recognition
Quickly divide by powers of 3$3$: 50/3 rightarrow 16$50/3 \rightarrow 16$; 16/3 rightarrow 5$16/3 \rightarrow 5$; 5/3 rightarrow 1$5/3 \rightarrow 1$. Summing them up yields 16 + 5 + 1 = 22$16 + 5 + 1 = 22$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q2025Permutation of Multiset
The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to:
A.360$360$
B.45$45$
C.2520$2520$
D.1820$1820$
Solution
### Related Formula
The number of permutations of n$n$ objects where p$p$ are of one kind, q$q$ are of another kind, and r$r$ are of a third kind is:
textTotal Permutations = fracn!p! cdot q! cdot r!$\text{Total Permutations} = \frac{n!}{p! \cdot q! \cdot r!}$
### Core Logic
The sequence has 10$10$ terms chosen from \0, 1, 2\$\{0, 1, 2\}$. It contains exactly five 1$1$s and exactly three 2$2$s. This leaves exactly 10 - 5 - 3 = 2$10 - 5 - 3 = 2$ terms to be filled by 0$0$s.
### Step 1: Arrangement Calculation
We need to arrange five 1$1$s, three 2$2$s, and two 0$0$s. The number of unique sequences is:
textTotal Sequences = frac10!5! cdot 3! cdot 2!$\text{Total Sequences} = \frac{10!}{5! \cdot 3! \cdot 2!}$textTotal Sequences = frac10 times 9 times 8 times 7 times 63 times 2 times 1 times 2 times 1 = 2520$\text{Total Sequences} = \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1} = 2520$
### Pattern Recognition
Note that sequences can start with 0$0$ since it asks for general sequences of ten terms rather than a standard non-zero multi-digit number representation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Permutations and Combinations
Q582025Geometry 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
Q572025Practical 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
More Permutations and Combinations Questions — jee_main_2025_24_jan_morning
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