Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The problem is equivalent to finding the number of integer solutions to the equation $x_1+x_2+x_3+x_4+x_5+x_6+x_7=12$ with $1 \le x_i \le 4$. By substituting $y_i = x_i - 1$, we convert this into $y_1+y_2+y_3+y_4+y_5+y_6+y_7=5$, where $0 \le y_i \le 3$. Using the stars and bars formula, the number of non-negative solutions is $\binom{5+7-1}{7-1} = 462$. We then subtract the cases violating the condition $y_i \le 3$, where at least one $y_i \ge 4$, which occurs in $7 \times \binom{1+7-1}{7-1} = 49$ instances, resulting in a final count of 413.

Divisibility by 15 requires a number to be divisible by both 3 and 5, which means the last digit must be 5 and the sum of all digits must be a multiple of 3. Given the digits {1, 2, 3, 5}, we seek 4-digit sequences d1d2d35 such that d1 + d2 + d3 + 5 is a multiple of 3. By testing combinations of these digits, we identify the sets {1, 1, 2}, {2, 2, 3}, {3, 3, 1}, {1, 1, 5}, {2, 3, 5}, and {3, 3, 5}, which yield 3, 3, 3, 3, 6, and 3 permutations respectively. Summing these possibilities gives a total of 21 unique numbers.

To determine the rank of a word in a dictionary, we arrange all the letters alphabetically and systematically count the number of words that precede the target word. The letters of MONDAY are A, D, M, N, O, Y. Words starting with A, D, M, N, and O are calculated using permutations of the remaining letters. Specifically, we calculate permutations for all possible prefixes alphabetically until reaching MONDAY: $5!$ for A, $5!$ for D, $4!$ for MA, $4!$ for MD, $4!$ for MN, $3!$ for MOA, $3!$ for MOD, and $2!$ for MONA, plus the word itself, yielding a total rank of 327.

Arrangements at a round table with adjacency restrictions are best approached by first seating the unrestricted individuals and then placing the restricted ones in the resulting gaps. Arranging 7 boys in a circle provides $6!$ ways to seat them, which creates 7 available spaces to place the 5 girls using ${}^{7}C_{5}$ selection methods. Multiplying these by the $5!$ ways to arrange the girls within those gaps leads to $6! \times {}^{7}C_{5} \times 5!$, which corresponds to $126 \times (5!)^2$.

The equation $x+y+z=21$ with constraints $x \ge 1, y \ge 3, z \ge 4$ is transformed into non-negative variables by setting $x'=x-1, y'=y-3, z'=z-4$. Substituting these into the original equation yields $x'+y'+z'=13$. The number of non-negative integral solutions for this form is calculated using the combination formula $\binom{n+r-1}{r-1}$, where $n=13$ and $r=3$, resulting in $\binom{13+3-1}{3-1} = \binom{15}{2}$, which equals 105.

To transport $8$ distinct persons into $3$ distinct cars with a capacity constraint of at most $3$ per car, we partition the $8$ persons into group sizes of $(3, 3, 2)$ in various orderings. The number of ways to assign these groups to $3$ distinct cars is calculated by identifying the distribution of capacities: $(3, 3, 2)$, $(3, 2, 3)$, and $(2, 3, 3)$. For each configuration, the number of ways is given by the multinomial coefficient inom{8}{3}inom{5}{3}inom{2}{2} = 560. Summing these $3$ configurations gives $560 + 560 + 560 = 1680$ ways.

Combinatorial selection based on fixed fruit quantities follows the multiplication principle for independent choices. We analyze the three valid cases for selecting 5 fruits with at least 2 oranges, 1 red apple, and 1 white apple: $2O, 2RA, 1WA$; $2O, 1RA, 2WA$; and $3O, 1RA, 1WA$. Calculating the combinations for each scenario results in ${}^{8}C_{2} \times {}^{7}C_{2} \times {}^{5}C_{1} = 2940$, ${}^{8}C_{2} \times {}^{7}C_{1} \times {}^{5}C_{2} = 1960$, and ${}^{8}C_{3} \times {}^{7}C_{1} \times {}^{5}C_{1} = 1960$, which sum to $6860$.

For a number to be odd, its last digit must be 1, 3, or 5. We calculate the arrangements for each case by fixing the last digit and permuting the remaining six digits, accounting for repeated values using the formula for permutations of a multiset. Fixing 1 or 5 as the last digit results in $\frac{6!}{3!2!} = 60$ arrangements each, while fixing 3 as the last digit results in $\frac{6!}{3!} = 120$ arrangements. Summing these counts results in $60 + 120 + 60 = 240$.

For a 4-digit ATM code with distinct digits from $\{0, 1, \dots, 7\}$ where the maximum digit is 7, the sum of the first two digits must equal the sum of the last two. By partitioning the available digits into pairs with the same sum $\alpha$ and ensuring 7 is included, we systematically count the valid combinations. Calculating the permutations for all possible sums that satisfy these conditions results in 72 possible codes.

This problem asks for the number of derangements of $5$ students, which is the number of permutations where no student occupies their original position, denoted by $D_n$. Using the formula D_n = n! imes ig( rac{1}{0!} - rac{1}{1!} + rac{1}{2!} - rac{1}{3!} + rac{1}{4!} - rac{1}{5!} ig), we substitute $n=5$ to get 120 imes ig( 1 - 1 + rac{1}{2} - rac{1}{6} + rac{1}{24} - rac{1}{120} ig). Simplifying the expression gives $60 - 20 + 5 - 1$, which results in $44$ ways.

In a descending arrangement of 5-digit numbers using digits {7, 5, 3, 2, 1}, we determine the rank by counting how many numbers are strictly greater than 35337. Numbers starting with 7 or 5 each contribute 5^4 = 625, totaling 1250. We then count numbers starting with 37 (125), followed by 357 and 355 (25 + 25 = 50), then 3537 and 3535 (5 + 5 = 10). Adding these results, we find 1250 + 125 + 50 + 10 = 1435 numbers appear before 35337, making its serial number 1436.

The number of ways to select 3 boys from $b$ and 2 girls from $g$ is given by the product $\binom{b}{3} \binom{g}{2} = 168$. Expanding the combinations results in $\frac{b(b-1)(b-2)}{6} \times \frac{g(g-1)}{2} = 168$, which simplifies to $b(b-1)(b-2) \times g(g-1) = 2016$. Factoring 2016 into consecutive integer products, we identify $8 \times 7 \times 6 = 336$ and $3 \times 2 = 6$, giving $b=8$ and $g=3$, which leads to $b+3g = 8 + 3(3) = 17$.

The total number of passwords of a length $n$ is $10^n$, as each character can be one of 5 letters or 5 numbers. A password has at least one number if it is not composed entirely of letters, meaning we subtract the count of letter-only passwords, $5^n$, from the total, $10^n$. Calculating this for lengths 6, 7, and 8 gives $(10^6 - 5^6) + (10^7 - 5^7) + (10^8 - 5^8)$, which simplifies to $5^6(2^6 + 2^7 \cdot 5 + 2^8 \cdot 25 - 1 - 5 - 25) = 5^6(64 + 640 + 6400 - 31) = 7073 \times 5^6$.

A four-digit number $abcd$ satisfies the criteria if each of the first three digits is a multiple of the last digit $d$. For $d=1$, any digit from 1 to 9 can be chosen for the first position and 0 to 9 for the others, leading to $9 \times 10 \times 10 = 900$ numbers. For each $d \in \{2, 3, 4, 5, 6, 7, 8, 9\}$, we count the multiples of $d$ available for $a$ (which cannot be 0), $b$, and $c$, which results in 100 numbers for $d=2$, 48 for $d=3$, 18 for $d=4$, 4 for $d=5$, and 16 combined for $d=6, 7, 8, 9$. Adding these values together gives a total of 1086 numbers.

For the product of the digits of a 5-digit number to be 36, the digits must be selected from the factors of 36 such that they form a 5-digit set, including 1s as fillers. We consider all combinations: {9, 4, 1, 1, 1}, {9, 2, 2, 1, 1}, {6, 6, 1, 1, 1}, {6, 3, 2, 1, 1}, {4, 3, 3, 1, 1}, and {3, 3, 2, 2, 1}. Calculating permutations for each case using $\frac{5!}{\text{repetition factors}}$ gives $20 + 30 + 10 + 60 + 30 + 30 = 180$.

Forming the group requires calculating the number of ways to select 3 boys and 3 girls while adhering to the constraint that $B_1$ and $B_2$ are not both present. The total ways to choose 3 boys from 10 is calculated as $\binom{10}{3} = 120$, and subtracting the 8 combinations where both $B_1$ and $B_2$ are selected leaves 112 valid choices for boys. Multiplying this by the $\binom{5}{3} = 10$ ways to choose the girls yields 1120.

Divisibility by 6 requires a number to be even and divisible by 3. Since the sum of the digits $1, 2, 3, 5, 6, 7$ is 24, a 5-digit number is divisible by 3 if the excluded digit is either 3 or 6. Excluding 3 leaves the set $1, 2, 5, 6, 7$, where even endings of 2 or 6 yield 48 permutations, and excluding 6 leaves $1, 2, 3, 5, 7$, where an ending of 2 yields 24 permutations, resulting in a total of 72.

The dictionary rank of a word is calculated by counting the number of permutations that appear alphabetically before it. With the letters {A, D, I, K, M, N, N}, there are $\frac{6!}{2!} = 360$ arrangements starting with each of A, D, I, and K, contributing 1440 words before we reach those starting with M. For the M-starting section, we systematically count permutations starting with MA-D, MA-I, and MA-K, each contributing $\frac{4!}{2!} = 12$ words, followed by further specific letter-by-letter placements that eventually sum to 1492.

The number of matrices is determined by identifying all sets of four entries from the set $0$ to $5$ that sum to a prime number $p$ where $2 < p < 8$. We examine the partitions of 3, 5, and 7 into four parts and calculate the permutations for each distinct set of entries. Summing the counts for all possible matrix configurations associated with these sums leads to the total of 180.

The score is determined by the equation $3x - 2y = 5$, where $x$ is the number of correct answers and $y$ is the number of incorrect answers, constrained by $x+y \leq 5$. Testing integer values, the only valid solution is $x=3$ and $y=2$. The number of ways is calculated by $\binom{5}{3} \times 1^3 \times 2^2$, which simplifies to $10 \times 4 = 40$.