Let S = p₁, p₂, …, p₁₀$S = \{p_1, p_2, \ldots, p_{10}\}$ be the set of the first ten prime numbers. Let A = S 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 ________.

Core Logic

First, analyze the frequency of letters in PQRPQRSTUVP. P: 3$P: 3$ Q: 2$Q: 2$ R: 2$R: 2$ S: 1$S: 1$ T: 1$T: 1$ U: 1$U: 1$ V: 1$V: 1$ Total types of distinct letters = 7$7$ (P, Q, R, S, T, U, V). We need to form 4-letter words. We break this down into mutually exclusive cases based on letter duplication.

Step 1: Case 1 - 3 Alike, 1 Different

Only 'P' can be chosen for the 3 alike letters. (1$1$ way: ¹C₁${^1C_1}$) The 1 different letter can be chosen from the remaining 6$6$ distinct letters. (⁶C₁${^6C_1}$) Arrangements = (4!)/(3!) = 4$\frac{4!}{3!} = 4$ Number of words = 1 × 6 × 4 = 24$1 \times 6 \times 4 = 24$

Step 2: Case 2 - 2 Alike, 2 Alike

We need to choose 2$2$ sets of letters that each appear at least twice. Letters eligible: P, Q, R (3 options). Choose 2 sets: ³C₂ = 3${^3C_2} = 3$ Arrangements = (4!)/(2!2!) = 6$\frac{4!}{2!2!} = 6$ Number of words = 3 × 6 = 18$3 \times 6 = 18$

Step 3: Case 3 - 2 Alike, 2 Different

Choose 1 set of letters to be alike from {P, Q, R}: ³C₁ = 3${^3C_1} = 3$ Choose 2 distinct letters from the remaining 6$6$ types: ⁶C₂ = 15${^6C_2} = 15$ Arrangements = (4!)/(2!) = 12$\frac{4!}{2!} = 12$ Number of words = 3 × 15 × 12 = 540$3 \times 15 \times 12 = 540$

Step 4: Case 4 - All 4 Different

Choose 4 distinct letters from the 7$7$ available types: ⁷C₄ = 35${^7C_4} = 35$ Arrangements = 4! = 24$4! = 24$ Number of words = 35 × 24 = 840$35 \times 24 = 840$

Step 5: Final Summation

Total words = Sum of all cases = 24 + 18 + 540 + 840 = 1422$24 + 18 + 540 + 840 = 1422$

Pattern Recognition

Letter grouping requires strict combinatorial partition cases (AAAB, AABB, AABC, ABCD). Never mix selection (ⁿCᵣ$^nC_r$) with arrangement (factorials) in the same unwritten step—do them rigorously sequentially.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Number of positive integral solutions for equation x₁ + x₂ + + xᵣ = n$x_1 + x_2 + \dots + x_r = n$ is ⁿ⁻¹Cᵣ₋₁$^{n-1}C_{r-1}$

Core Logic

Assuming oranges are identical, let x₁, x₂, x₃, x₄$x_1, x_2, x_3, x_4$ be the number of oranges given to the 4 children.

where x₁, x₂, x₃, x₄ ≥ 1$x_1, x_2, x_3, x_4 \geq 1$.

Step 1: Shift Variables

Let xᵢ = xᵢ' + 1$x_i = x_i' + 1$ such that xᵢ' ≥ 0$x_i' \geq 0$.

The number of non-negative integral solutions is given by n+r-1Cᵣ₋₁$^{n+r-1}C_{r-1}$, where n = 12, r = 4$n = 12, r = 4$.

Step 2: Calculate Combinations

Pattern Recognition

Standard "stars and bars" problem. "At least one" means we use ⁿ⁻¹Cᵣ₋₁$^{n-1}C_{r-1}$. Here ¹⁶⁻¹C₄₋₁ = ¹⁵C₃$^{16-1}C_{4-1} = ^{15}C_3$ directly.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Any subset of size r$r$ chosen from a set of n$n$ distinct numbers can be arranged in strictly descending order in exactly 1 way. Hence, combinations = arrangements.

Core Logic

For Set S$S$: We need 4 digits a, b, c, d$a, b, c, d$ such that a > b > c > d$a > b > c > d$. Since the numbers are strictly descending, d$d$ can be 0$0$ (e.g., 3210$3210$ is a valid 4-digit number). We simply need to choose 4 distinct digits from the 10 available digits 0, 1, 2, , 9$\{0, 1, 2, \dots, 9\}$. Once chosen, they can be arranged in decreasing order in exactly one unique way.

Step 1: Generating 5-Digit Multipliers

For Set P$P$: We need 5 digits whose product is 20. None of the digits can be 0. Digits must be from 1, 2, 3, , 9$\{1, 2, 3, \dots, 9\}$. The prime factorization of 20 is 2² × 5$2^2 \times 5$. The possible sets of 5 digits that multiply to 20 are: Case 1: 5, 4, 1, 1, 1$\{5, 4, 1, 1, 1\}$ Case 2: 5, 2, 2, 1, 1$\{5, 2, 2, 1, 1\}$

Step 2: Counting Permutations

Now calculate the number of permutations for each case to form 5-digit numbers: For Case 1 5, 4, 1, 1, 1$\{5, 4, 1, 1, 1\}$:

For Case 2 5, 2, 2, 1, 1$\{5, 2, 2, 1, 1\}$:

Total for Set P$P$:

Step 3: Final Sum

Total sum n(S) + n(P) = 210 + 50 = 260$n(S) + n(P) = 210 + 50 = 260$.

Pattern Recognition

A strictly monotonic digits condition always simplifies a permutation problem into a pure combination selection (ⁿCᵣ$^nC_r$). Product conditions require basic integer partition mapping and multiset permutation formulas.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Divisibility by 3: sum of digits must be a multiple of 3. Numbers must be 4 digits, starting with 5 (since > 5000$> 5000$ and < 9000$< 9000$, the first digit must be 5 as 9 is out of bounds for the first digit).

Core Logic

Format: 5 _ _ _$5 _ _ _$ Sum of digits = 5 + x₂ + x₃ + x₄ = 3k$= 5 + x_2 + x_3 + x_4 = 3k$. Available digits to pick from: 0, 1, 2, 5, 9$0, 1, 2, 5, 9$. (Their mod 3 values are 0, 1, 2, 2, 0$0, 1, 2, 2, 0$). Instead of full manual listing, one can catalog combinations summing to 3k$3k$ when added to 5.

Step 1: Analyzing Cases

Sum x₂+x₃+x₄$x_2+x_3+x_4$ must be ≡ 1 3$\equiv 1 \pmod 3$ since 5 ≡ 2 3$5 \equiv 2 \pmod 3$. Categorizing remaining 3 slots by combinations: (1) All different elements (from 0, 1, 9, 2, 5 - wait, 5 is fixed but can repeat, we just need the 3 remaining slots). Let's trace valid combinations of 3 digits from 0, 1, 2, 5, 9$0, 1, 2, 5, 9$ summing to a required modulo. Combinations mapped manually matching modulo conditions: 0, 1, 9 ⇒ 5,0,1,9 ⇒ (3!)/(1!) = 6$0, 1, 9 \Rightarrow 5,0,1,9 \Rightarrow \frac{3!}{1!} = 6$ ways

(2) Two alike, two different: 0, 0, 1 ⇒ 5, 0, 0, 1 ⇒ (3!)/(2!) = 3$0, 0, 1 \Rightarrow 5, 0, 0, 1 \Rightarrow \frac{3!}{2!} = 3$ ways 1, 1, 2 ⇒ 5, 1, 1, 2 ⇒ 3$1, 1, 2 \Rightarrow 5, 1, 1, 2 \Rightarrow 3$ ways 2, 2, 0 ⇒ 5, 2, 2, 0 ⇒ 3$2, 2, 0 \Rightarrow 5, 2, 2, 0 \Rightarrow 3$ ways 2, 2, 9 ⇒ 5, 2, 2, 9 ⇒ 3$2, 2, 9 \Rightarrow 5, 2, 2, 9 \Rightarrow 3$ ways 5, 0, 2 ⇒ 5, 5, 0, 2 ⇒ 3! = 6$5, 0, 2 \Rightarrow 5, 5, 0, 2 \Rightarrow 3! = 6$ ways 5, 2, 9 ⇒ 5, 5, 2, 9 ⇒ 3! = 6$5, 2, 9 \Rightarrow 5, 5, 2, 9 \Rightarrow 3! = 6$ ways 1, 9, 9 ⇒ 5, 1, 9, 9 ⇒ 3$1, 9, 9 \Rightarrow 5, 1, 9, 9 \Rightarrow 3$ ways

(3) Three alike (and 5): 5, 5, 0 ⇒ 5, 5, 5, 0 ⇒ 3$5, 5, 0 \Rightarrow 5, 5, 5, 0 \Rightarrow 3$ ways 5, 5, 9 ⇒ 5, 5, 5, 9 ⇒ 3$5, 5, 9 \Rightarrow 5, 5, 5, 9 \Rightarrow 3$ ways

(4) Two alike and Two other alike (including the leading 5): 5, 1, 1 ⇒ 5, 5, 1, 1 ⇒ (3!)/(2!) = 3$5, 1, 1 \Rightarrow 5, 5, 1, 1 \Rightarrow \frac{3!}{2!} = 3$ ways

Step 2: Total Sum

Total permutations = 6 + (3+3+3+3+6+6+3) + (3+3) + 3$= 6 + (3+3+3+3+6+6+3) + (3+3) + 3$ = 6 + 27 + 6 + 3 = 42$= 6 + 27 + 6 + 3 = 42$

Pattern Recognition

Breaking the 4-digit number into a fixed constraint (5$5$ at thousands place) and tracking permutations of the remaining 3 slots drastically reduces the combinatorial tree. Grouping by likeness handles identical digits flawlessly.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Core Logic

Alphabetical order of letters in "UDAYPUR": A, D, I (Wait, no I. A, D, P, R, U, U, Y). Let's list properly: A, D, P, R, U, U, Y.

Step 1: Words Starting with Previous Letters

Start fixing the first letter from the sorted array:

Words starting with A arrow (6!)/(2!) = (720)/(2) = 360$A \rightarrow \frac{6!}{2!} = \frac{720}{2} = 360$ Words starting with D arrow (6!)/(2!) = 360$D \rightarrow \frac{6!}{2!} = 360$ Words starting with P arrow (6!)/(2!) = 360$P \rightarrow \frac{6!}{2!} = 360$ Words starting with R arrow (6!)/(2!) = 360$R \rightarrow \frac{6!}{2!} = 360$

Step 2: Words starting with U

Now we reach U$U$. Fix U$U$ and move to the second letter. Available letters: A, D, P, R, U, Y.

Words starting with UA arrow 5! = 120$UA \rightarrow 5! = 120$ (Since one U is fixed, no repetition left)

Next prefix is UD$UD$. Fix UD$UD$, move to the third letter. Available: A, P, R, U, Y. A$A$ is the target, so fix A$A$. We are at UDA$UDA$. Move to fourth letter. Available: P, R, U, Y. Y$Y$ is target.

Words starting with UDAP arrow 3! = 6$UDAP \rightarrow 3! = 6$ Words starting with UDAR arrow 3! = 6$UDAR \rightarrow 3! = 6$ Words starting with UDAU arrow 3! = 6$UDAU \rightarrow 3! = 6$

Next is UDAY$UDAY$. Fix UDAY$UDAY$. Available: P, R, U.

Step 3: Final Count

Alphabetical order for remaining three letters is P, R, U. The target word is UDAYPUR.

Words starting with UDAYPRU arrow 1$UDAYPRU \rightarrow 1$ (since U is after R) Next word is UDAYPUR arrow 1$UDAYPUR \rightarrow 1$

Total Rank = 360 + 360 + 360 + 360 + 120 + 6 + 6 + 6 + 1 + 1 = 1580$360 + 360 + 360 + 360 + 120 + 6 + 6 + 6 + 1 + 1 = 1580$.

Pattern Recognition

When repeating letters exist (like U in UDAYPUR), dividing by the factorial of repeated characters is critical for preceding branches. Once one of the repeating letters is consumed in the fixed prefix (e.g. at UA), the remaining pool has unique letters, dropping the denominator.

Chapter Mix

Class 11 Maths: Permutations and Combinations