Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The digits of 123412341 consist of three 1s and two each of 2, 3, and 4. The 4 even positions require arranging $\{2,2,4,4\}$, which yields $\frac{4!}{2!2!} = 6$ arrangements, while the 5 odd positions require arranging the remaining digits $\{1,1,1,3,3\}$, which yields $\frac{5!}{3!2!} = 10$ arrangements. Multiplying these independent counts gives a total of $6 \times 10 = 60$.

The placement principle for numbers between 5000 and 10000 dictates that the first digit must be one of the three options $5, 7,$ or $9$ to ensure the value exceeds 5000. For each choice, the remaining three digits are selected and arranged from the remaining four digits in the set $1, 3, 5, 7, 9$ without repetition. This results in $3 \times 4 \times 3 \times 2 = 72$ total numbers.

The condition that the greatest common divisor of a 4-digit number N and 54 is 2 implies that N must be divisible by 2 but not by 3, since 54 = 2 * 3^3. The total number of 4-digit integers divisible by 2 is 4500, while those divisible by 6 (which are divisible by both 2 and 3) is 1500. By subtracting the latter from the former, we determine that 4500 - 1500 = 3000 numbers satisfy the given condition.

The rank of a word in a dictionary is determined by calculating the number of permutations that precede it alphabetically. With the letters M, A, T, H, S, the sorted alphabetical order is A, H, M, S, T. For the word THAMS, we first count all permutations starting with A, H, M, or S, each contributing $4! = 24$ arrangements, summing to $96$. Next, we count permutations starting with T, specifically those starting with TA, where the remaining letters H, M, S can be ordered in $3!$ ways, but specifically looking for TAH..., TAM..., and TAS..., each contributing $2! = 2$ arrangements, totaling $6$ additional words. Thus, the $102$ words precede THAMS, making it the $103^{rd}$ word.

The number of matrices is determined by the coefficient of $x^p$ in the expansion of $(1+x+x^{2+}x^{3+}x^4)^4$, where the sum $S$ equals a prime number $p \in \{3, 5, 7, 11\}$. By expanding this polynomial, we find that the counts for sums 3, 5, 7, and 11 are 20, 52, 80, and 52 respectively. Summing these possibilities gives a total of 204 matrices.

Using the principle of inclusion-exclusion, we identify the counts of 3-digit numbers divisible by 2, 3, and 7, which are 450, 300, and 128 respectively. The number of 3-digit numbers divisible by 2 or 3 is $450 + 300 - 150 = 600$, while those divisible by 2 or 3 and also by 7 is calculated as $|(A \cap C) \cup (B \cap C)| = 64 + 43 - 21 = 86$. Subtracting these overlapping values from the union leaves $600 - 86 = 514$.

The identity $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$ is key to simplifying sums of combinations. We express the given series as $\sum_{k=45}^{51} {}^{k}C_{3}$. By repeatedly applying the property ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$, we can observe that adding and subtracting ${}^{45}C_{4}$ allows the sum to telescope into the form ${}^{52}C_{4} - {}^{45}C_{4}$.

To satisfy the condition that $a+b$ leaves a remainder of 2 when divided by 23, we identify all sums of the form $23k+2$ that are odd, as the sum of an even number $a$ and an odd number $b$ must be odd. This restricts the potential sums to $25, 71, 117$, and $163$, and by calculating the valid number of pairs $(a,b)$ for each sum within the defined ranges, we sum $12, 35, 42,$ and $19$ to get 108.

A mixed doubles match between couples requires selecting two couples from the total of n available couples and then forming a valid pair. The total number of ways to conduct such matches is given by the equation ${}^{n}C_{2} \times {}^{n-2}C_{2} \times 2 = 840$, which simplifies to $n(n-1)(n-2)(n-3) = 1680$. Since $8 × 7 × 6 × 5 = 1680$, it follows that $n=8$, representing 8 couples, which implies a total of 16 persons.

The series $\frac{1}{1!50!} + \frac{1}{3!48!} + \dots + \frac{1}{51!0!}$ is related to binomial coefficients by multiplying each term by $51!$, which transforms the sum into $\binom{51}{1} + \binom{51}{3} + \dots + \binom{51}{51}$. The sum of odd terms in the binomial expansion of $(1+1)^{51}$ equals $2^{51-1}$, so the series sums to $2^{50}$. Dividing by the factor $51!$ restores the original value to $\frac{2^{50}}{51!}$.

When arranging letters with restricted vowel placement, treat all vowels as a single composite unit to ensure they remain grouped. In the word INDEPENDENCE, there are 7 consonants and one vowel block (consisting of I, E, E, E, E), resulting in 8 total units to arrange while accounting for the repetitions of N, D, and the vowels. The calculation $\frac{8!}{3!2!} \times \frac{5!}{4!}$ determines the total number of valid arrangements, which simplifies to $3360 \times 5 = 16800$.

The sum of the digits must be divisible by 3 and the last digit must be 4 for the number to be divisible by 6. This requires the sum of the first 5 digits, $S'$, to satisfy $S' + 4 \equiv 0 \pmod 3$, or $S' \equiv 2 \pmod 3$. In sequences of length $n$ using the digits {4, 5, 9}, the total count is $3^n$, and exactly one-third of these will have a sum congruent to any value modulo 3, resulting in $3^4 = 81$ such numbers.

To form a nine-digit number using three digits in an arithmetic progression, each used thrice, we calculate the total arrangements and apply combinatorial constraints. Given the digits must contain at least one consecutive triplet in arithmetic progression, we account for the specific placements of these triplets within the nine-digit sequence. By calculating the permutations of the digits while fixing the consecutive structure required for the arithmetic progression, we arrive at 1260 distinct numbers.

Dictionary arrangement relies on counting all possible permutations of the letters {G, H, O, T, U} that precede the target word TOUGH in alphabetical order. Words starting with G, H, or O contribute 3 * 4! = 72 entries. Continuing with words starting with T, we count those beginning with TG (3! = 6) and TH (3! = 6), followed by those starting with TO, specifically TOG (2! = 2) and TOH (2! = 2). Finally, adding the sequence starting with TOU, we find the word TOUGH is the 72 + 6 + 6 + 2 + 2 + 1 = 89th word.

The serial number depends on counting all five-digit numbers made of the given digits {0,2,3,4,7,9} that precede 42923 in ascending order. Numbers starting with 2 or 3 contribute $2 \times 6^4 = 2592$ values. Further systematically counting the possibilities for numbers starting with 40, 41, and then those within the 42xxx range accounts for the remaining increments, leading to the serial number 2997.

The total number of elements in the Cartesian product $A \times B$ is $5 \times 2 = 10$. The number of subsets with size $r$ is given by the binomial coefficient $\binom{10}{r}$, so we calculate the sum $\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6}$. This evaluates to $120 + 210 + 252 + 210 = 792$.

A number is divisible by 6 if it is even and the sum of its digits is a multiple of 3. Given the digits 1, 2, 3, 4, and 5 with repetition, we fix the last digit as 2 or 4 to ensure the number is even. For each fixed last digit, we identify all pairs of the first two digits such that the total sum of the three digits is divisible by 3. Summing these valid combinations for both ending digits, 2 and 4, results in 8 possibilities each, for a total of 16.

Distributing 20 distinct items into 3 distinct bins such that no bin is empty is an application of the inclusion-exclusion principle, represented by $3^{20} - \binom{3}{1}2^{20} + \binom{3}{2}1^{20}$. Substituting the values yields $3486784401 - 3 \times 1048576 + 3$, which simplifies to $3483638676$.

Forming integers greater than 7000 using digits $\{3,5,6,7,8\}$ without repetition requires counting 4-digit and 5-digit numbers separately. For 4-digit numbers starting with 7 or 8, there are $2 \times (4 \times 3 \times 2) = 48$ possibilities. Since all 5-digit numbers are greater than 7000, there are $5! = 120$ such numbers. The total sum is $48 + 120 = 168$.

Alphabetize the letters as B, C, I, L, P, U. Words starting with B, C, I, and L each have $5! = 120$ variations, totaling $480$. For words starting with P, count the permutations starting with PB, PC, PI, and PL, each providing $4! = 24$ variations for a subtotal of $96$. Finally, include the specific sequences starting with PU, namely PUBC, PUBI, PUBLC, and PUBLIC, which contribute $2 + 2 + 1 + 1 = 6$ additional words for a total serial number of $582$.