Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The problem of finding the number of $3$-digit integers whose digits sum to $10$ can be modeled as finding the number of non-negative integral solutions to $x + y + z = 10$, subject to the constraints $1 \le x \le 9$ and $0 \le y, z \le 9$. By substituting $T = x - 1$ where $0 \le T \le 8$, the equation transforms into $T + y + z = 9$, allowing us to use the stars and bars method $\binom{n+k-1}{k-1}$. Calculating $\binom{9+3-1}{3-1}$ yields $55$ combinations, and after subtracting the single invalid case where $T = 9$ (which would force $x = 10$), the total number of valid integers is $54$.

To select 5 questions while ensuring at least one from each of the three sections, we can have distributions of either $(1, 2, 2)$ or $(1, 1, 3)$ questions per section. For the $(1, 2, 2)$ distribution, there are $\binom{5}{1} \binom{5}{2} \binom{5}{2} = 500$ ways, and since any of the three sections could be the one with only 1 question, we have $3 \times 500 = 1500$ ways. For the $(1, 1, 3)$ distribution, there are $\binom{5}{1} \binom{5}{1} \binom{5}{3} = 250$ ways, and similarly, there are $3 \times 250 = 750$ ways. Summing these yields $1500 + 750 = 2250$.

The maximum value of a binomial coefficient ${}^{n}C_{r}$ occurs at the middle index, specifically at $\frac{n}{2}$ for even $n$ and $\frac{n \pm 1}{2}$ for odd $n$. Here $a = {}^{19}C_{10}$, $b = {}^{20}C_{10}$, and $c = {}^{21}C_{10}$. Using the identity ${}^{n}C_{r} = \frac{n}{r} \times {}^{n-1}C_{r-1}$, we find $b = 2a$ and $c = \frac{21}{11}b = \frac{42}{11}a$. Expressing the ratio $a:b:c$ results in $1:2:\frac{42}{11}$, which simplifies to $11:22:42$.

The binomial distribution formula $\binom{n}{r} \times a^r \times b^{n-r}$ calculates the number of ways to arrange $r$ successes and $n-r$ failures, where $a$ and $b$ represent the number of choices for each. With $6$ questions and $4$ correct answers, we use $\binom{6}{4}$ to select the successful questions, each having $1$ correct answer. Since each of the $2$ incorrect questions has $4 - 1 = 3$ alternative options, the total number of ways is calculated as $\binom{6}{4} \times 1^4 \times 3^2$, which simplifies to $15 \times 9$, resulting in $135$.

Treating each family as a single entity ensures members of the same family stay together, allowing the three family groups to be arranged in $3!$ ways. Within their respective groups, the members of the three families can be arranged in $3!$, $3!$, and $4!$ ways. Multiplying these independent arrangements results in the total number of ways given by $3! \times 3! \times 3! \times 4!$, which simplifies to $(3!)^3 \cdot 4!$.

A 6-digit number composed of exactly five distinct digits involves one digit repeating exactly once. The number of such configurations is found by choosing the repeated digit in $\binom{5}{1}$ ways and then arranging the 6 items, which includes two identical digits, in $\frac{6!}{2!}$ ways, leading to the calculation $\binom{5}{1} \times \frac{6!}{2!}$.

The lexicographical position of a word is determined by calculating the number of permutations that precede it when letters are arranged in alphabetical order. For the word MOTHER, the letters are E, H, M, O, R, T, and we count permutations starting with letters that appear before M (E and H), which gives $2 \times 5!$ words. Continuing systematically by fixing M and counting the permutations for words starting with ME, MH, MOE, MOH, MOR, and MOTE, we sum $2 \times 4! + 3 \times 3! + 2!$ and finally add $1$ to account for the word MOTHER itself, resulting in a total rank of $309$.

Selecting items from a group depends on whether the items are considered distinct, using combinations, or identical, using partitions. Assuming distinct marbles, we subtract the forbidden scenario of all 4 being red from the total combinations: ${}^{12}C_{4} - {}^{5}C_{4} = 495 - 5 = 490$. If the marbles are identical, we find the number of ways to satisfy $r+b+w=4$ given $r \le 3$, $b \le 4$, and $w \le 3$, which yields 13 possible combinations of marble counts.

The number of arrangements for a restricted permutation is found by fixing the specific position and accounting for remaining digit constraints. With the tens place occupied by 2, we must choose four digits from the remaining nine, ensuring the ten-thousands place is not zero. This results in 8 options for the ten-thousands place and $8 \times 7 \times 6$ ways to fill the remaining three positions, giving a total of $8 \times 8 \times 7 \times 6 = 2688$ numbers. Setting $336k = 2688$ gives $k = 8$.

The gap method ensures that vowels are kept apart by arranging the consonants first and then placing the vowels into the spaces between them. With the consonants L, T, T, and R, the number of distinct arrangements is $\frac{4!}{2!} = 12$. Since there are 5 available gaps created by these 4 consonants, the 2 identical vowels can be placed in ${}^{5}C_{2} = 10$ ways. Multiplying these counts gives $12 \times 10 = 120$ possible arrangements.

Permutations of multiset items require categorization based on the presence of duplicate elements to account for distinct arrangements. For the word EXAMINATION, we have 8 distinct letters with three pairs of duplicates: N, A, and I. Calculating the cases for 4-letter words, two pairs yield ${}^{3}C_{2} \times \frac{4!}{2!2!} = 18$ permutations, one pair with two distinct letters yields ${}^{3}C_{1} \times {}^{7}C_{2} \times \frac{4!}{2!} = 756$ permutations, and four distinct letters yield ${}^{8}C_{4} \times 4! = 1680$ permutations. Summing these provides the total of 2454.

Recognizing that the product of an integer $n$ and its permutation $P_{n-1}$ is equivalent to $n!$, the first expression simplifies to the series $2! - 3! + 4! - \dots + 52!$. Combining this with the second series $1! - 2! + 3! - 4! + \dots + 51!$ creates a telescoping sum where most terms cancel each other out. After all intermediate terms are eliminated, only the initial $1!$ and the final $52!$ remain, resulting in a total value of $1! + 52!$.

In a circular arrangement of $n$ stations, the blue lines represent the polygon's sides, totaling $n$ connections, while the red lines represent all possible diagonals calculated as $\frac{n(n-1)}{2} - n$. Given that the number of red lines is $99$ times the number of blue lines, we establish the equation $\frac{n(n-1)}{2} - n = 99n$. Dividing both sides by $n$ for $n > 2$ simplifies the expression to $\frac{n-1}{2} - 1 = 99$, which solves to $\frac{n-1}{2} = 100$, yielding $n = 201$.

The selection of a committee is calculated by summing the combinations for each valid gender distribution. For at least 6 males, the sum is $\binom{8}{6}\binom{5}{5} + \binom{8}{7}\binom{5}{4} + \binom{8}{8}\binom{5}{3} = 28 + 40 + 10 = 78$. For at least 3 females, the sum is $\binom{5}{3}\binom{8}{8} + \binom{5}{4}\binom{8}{7} + \binom{5}{5}\binom{8}{6} = 10 + 40 + 28 = 78$. Both conditions result in 78 total ways, making them equal.

Selecting a team of 3 students from 5 boys and $n$ girls, where the team must include at least one boy and one girl, corresponds to the combinations of (1 boy, 2 girls) or (2 boys, 1 girl). This is expressed by the equation $\binom{5}{1} \binom{n}{2} + \binom{5}{2} \binom{n}{1} = 1750$, which simplifies to the quadratic $n^2 + 3n - 700 = 0$, yielding $n=25$ as the positive integer solution.

Connecting the tops of 20 pillars on a circular stadium with beams between non-adjacent pillars is equivalent to drawing all possible chords between 20 points and subtracting the edges of the 20-sided polygon. The number of total possible lines is $\binom{20}{2}$, and subtracting the 20 edges gives the result $\binom{20}{2} - 20 = 190 - 20 = 170$.

The number of natural numbers less than 7000 is determined by summing the possibilities for each digit length using the set {0, 1, 3, 7, 9}. For 1, 2, or 3-digit numbers, we exclude zero from the leading position to form $4 + 20 + 100 = 124$ combinations, while for 4-digit numbers, the leading digit must be 1 or 3 to remain below 7000, yielding $2 \times 5 \times 5 \times 5 = 250$ additional combinations. Summing these provides a total of 374 natural numbers.

There are four even-numbered positions available for the three odd digits (1, 1, 3), which can be selected in $\binom{4}{3}$ ways and arranged in $\frac{3!}{2!}$ ways. The remaining six even digits (2, 2, 2, 2, 4, 4) fill the six vacant positions in $\frac{6!}{4!2!}$ ways. Multiplying these counts gives $4 \times 3 \times 15 = 180$.

Choosing 10 objects from 31, where 10 are identical and 21 are distinct, requires selecting $r$ items from the distinct set and $10-r$ from the identical set. The total number of ways is calculated as the sum $\sum_{r=0}^{10} \binom{21}{r}$. By the symmetry property of binomial coefficients $\binom{n}{r} = \binom{n}{n-r}$, this sum simplifies to half of $2^{21}$, resulting in $2^{20}$.

To find the count of four-digit numbers strictly greater than 4321 using the digits 0 through 5 with repetition, we evaluate disjoint ranges. The count includes 4 values for 4322-4325, $3 \times 6 = 18$ values for 4330-4355, $1 \times 6 \times 6 = 36$ values for 4400-4455, $1 \times 6 \times 6 = 36$ values for 4500-4555, and $1 \times 6^3 = 216$ values for 5000-5555. Summing these yields $4 + 18 + 36 + 36 + 216 = 310$.