Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The rank is determined by listing all valid permutations alphabetically while excluding any word with consecutive Rs. Counting the permutations that begin with A and E yields $36 + 36 = 72$. Adding the counts for words starting with FAE, FAM, FARE, and the word FARMER itself, we obtain $72 + 1 + 1 + 2 + 1 = 77$.

Combinatorial selection problems involving multiple constraints can be solved by partitioning the problem into mutually exclusive cases based on the number of students selected from each category. Given the requirements of at least two students from each class and a total of ten students, with the additional constraint that the combined count from class 10 and class 11 does not exceed five, the valid distributions for $(c_{10}, c_{11}, c_{12})$ are $(2, 2, 6)$, $(2, 3, 5)$, and $(3, 2, 5)$. Calculating the number of ways for these cases using combinations, we find ${}^{5}C_{2} \cdot {}^{6}C_{2} \cdot {}^{8}C_{6} = 4200$, ${}^{5}C_{2} \cdot {}^{6}C_{3} \cdot {}^{8}C_{5} = 11200$, and ${}^{5}C_{3} \cdot {}^{6}C_{2} \cdot {}^{8}C_{5} = 8400$. Summing these results yields a total of $23800$, which implies $k = 238$.

The sum of all numbers formed by a set of digits is calculated by finding the sum of digits in each place value across all possible arrangements. For the digits {1, 2, 2, 3}, there are $\frac{4!}{2!} = 12$ distinct numbers. In any specific place, 1 occurs 3 times, 2 occurs 6 times, and 3 occurs 3 times. The sum of the digits in any single column is $1(3) + 2(6) + 3(3) = 24$. Multiplying this sum by the value of the positions $(10^3 + 10^2 + 10^1 + 10^0 = 1111)$ gives $24 \times 1111 = 26664$.

The formation of seven-digit integers using digits {1, 2, 3} with a fixed sum of digits requires partitioning the number into specific counts of each digit. By solving the system of equations where the total number of digits is 7 and their sum is 10, we find two possible compositions: four 1s and three 2s or five 1s, one 2, and one 3. The total count is calculated by summing the permutations for each case: $\frac{7!}{4!3!} + \frac{7!}{5!1!1!} = 35 + 42 = 77.$

To find the number of arrangements where all consonants never appear together, we subtract the arrangements where they do form a single block from the total permutations. Total permutations for the six letters are $6! = 720$. Treating the four consonants as a single unit, there are $3! \times 4! = 144$ prohibited arrangements, resulting in $720 - 144 = 576$ valid permutations.

The total number of distinct triangles formed by choosing 3 points from 15 is $\binom{15}{3} = 455$. We subtract the triangles where the sum of the vertex indices $i+j+k=15$. Listing these sets $(i,j,k)$ where $i<j<k$ reveals 12 such cases. Subtracting these gives $455 - 12 = 443$ valid triangles.

To form a triangle, three points must be selected such that they are not collinear, which happens by choosing points from different sides of the triangle. The total number of valid triangles includes combinations of one point from each of the three sides-$3 imes 5 imes 6 = 90$-plus combinations where two points are selected from one side and one point from a different side. Summing these cases-$\binom{3}{2} \times 5$, $\binom{3}{2} \times 6$, $\binom{5}{2} \times 3$, $\binom{5}{2} \times 6$, $\binom{6}{2} \times 3$, and $\binom{6}{2} \times 5$-results in $15 + 18 + 30 + 60 + 45 + 75 = 243$. Adding the 90 cases where one point is picked from each side yields a total of 333 triangles.

Forming a number greater than 10,000 using five distinct digits chosen from the set {0, 2, 4, 6, 8} requires constructing a five-digit number. Because the first digit cannot be zero, there are four possible choices for the ten-thousands place. For the remaining four positions, the digits are chosen without replacement from the remaining set, which now includes zero, resulting in a permutation of four remaining digits. Multiplying the choices for each position gives $4 \times 4 \times 3 \times 2 \times 1 = 96$ distinct numbers.

The given expression can be simplified by rewriting the term $(r^3 + 6r^2 + 2r + 5)$ as $(r+1)(r+2)(r+3) - 9(r+1) + 8$. Substituting this into the summation, the general term becomes $r![(r+1)(r+2)(r+3)] - 9r!(r+1) + 8r!$, which simplifies to $(r+3)! - (r+1)! - 8((r+1)! - r!)$. Applying the sum to this telescoping series from 1 to 10 results in $(13! + 12! - 3! - 2!) - 8(11! - 1!)$. Evaluating the factorials and simplifying the expression leads to $(156 \cdot 11! + 12 \cdot 11! - 8 \cdot 11!)$, which equates to $160 \cdot 11!$, confirming that $\alpha = 160$.

The summation revolves around the fundamental definition of permutations where ${}^n P_n$ simplifies to $n!$, turning the expression into the sum of $r \cdot r!$ for $r$ from $1$ to $15$. A powerful technique here is to rewrite $r \cdot r!$ as $(r+1-1) \cdot r!$, which expands to $(r+1)! - r!$. This creates a telescoping series where terms cancel sequentially, effectively reducing the entire sum to the final term minus the initial term, resulting in $16! - 1$.

Comparing $16! - 1$ to the provided form ${}^{q}P_{r} - s$, we determine that $q = 16$, $r = 16$, and $s = 1$. Substituting these values into the required term ${}^{q+s}C_{r-s}$ leads to ${}^{17}C_{15}$. Because the combination formula follows the identity ${}^n C_k = {}^n C_{n-k}$, this is equivalent to ${}^{17}C_{2}$. Computing $\frac{17 \times 16}{2}$ results in 136.

Determining the count of three-digit numbers formed from {1, 2, 3, 4, 5} divisible by 3 or 5 involves inclusion-exclusion based on divisibility rules. Numbers divisible by 3 must have a digit sum divisible by 3, yielding 24 valid permutations, while those divisible by 5 must end in 5, yielding 12 permutations. Subtracting the intersection of these sets, which are numbers divisible by 15, gives the final total of $24 + 12 - 4 = 32.$

A six-digit palindrome takes the form $xyzzyx$, where $x$, $y$, and $z$ are digits. For this number to be divisible by 55, it must be divisible by both 5 and 11. Divisibility by 5 requires the last digit $x$ to be 5, as the leading digit cannot be zero. Divisibility by 11 requires the alternating sum of digits $(x+z+y) - (y+z+x)$ to be a multiple of 11, which simplifies to 0 and is inherently satisfied for any values of $y$ and $z$. Since $x$ is fixed as 5 and both $y$ and $z$ can independently take any of the 10 possible digits from 0 to 9, there are $10 \times 10 = 100$ such palindromes.

Forming a scientific committee with the condition that there are at least 2 Indians and exactly twice as many foreigners requires checking the possible committee compositions. We evaluate the cases (2 Indians, 4 foreigners), (3 Indians, 6 foreigners), and (4 Indians, 8 foreigners) using combinations to select members from the available pools. Summing these independent possibilities gives $\binom{6}{2}\binom{8}{4} + \binom{6}{3}\binom{8}{6} + \binom{6}{4}\binom{8}{8} = 1050 + 560 + 15 = 1625.$

Consider the integers from 000 to 999, representing each as a three-digit sequence. This set contains 1000 numbers, creating a total of 3000 digit positions. Because each of the ten digits appears with equal frequency across all these positions, the digit 3 occurs $3000 / 10 = 300$ times. Since the number 1000 contains no 3s, the count remains 300.

Finding the number of positive integral solutions for the product $xyz = 24$ relies on the prime factorization of 24, which is $2^3 \cdot 3^1$. By expressing $x, y,$ and $z$ as products of these prime factors, we distribute the three factors of 2 and one factor of 3 among the three variables using the stars and bars formula, $\binom{n+k-1}{k-1}$. The total number of solutions is the product of the independent distributions, resulting in $\binom{3+3-1}{3-1} \times \binom{1+3-1}{3-1} = 10 \times 3 = 30.$

To form a three-digit even number without repeating digits from the set $\{0, 1, 3, 4, 6, 7\}$, the unit digit must be even. When the unit digit is 0, the remaining two positions can be filled by any of the 5 remaining digits in $5 \times 4 = 20$ ways. When the unit digit is either 4 or 6, there are 2 choices for the units place, 4 choices for the hundreds place (as zero cannot be placed there and the unit digit is already used), and 4 choices for the tens place, resulting in $2 \times 4 \times 4 = 32$ ways. Adding these cases together provides a total of 52 such numbers.

The total number of matches is the sum of all possible boy-versus-boy matches and girl-versus-girl matches between the two teams. With 7 boys in Team A and 4 boys in Team B, the number of boy matches is $7 \times 4 = 28$. Given there are $n$ girls in Team A and 6 girls in Team B, the girl matches count is $n \times 6$. Setting the sum $28 + 6n$ equal to the total of 52 matches leads to $6n = 24$, confirming $n = 4$.

The word SYLLABUS contains two pairs of identical letters (S, S and L, L) and four distinct letters (Y, A, B, U). To form a 4-letter word with two alike and two distinct, we first select one of the two pairs in ${}^{2}C_{1} = 2$ ways, then choose two distinct letters from the remaining five letters in ${}^{5}C_{2} = 10$ ways. Arranging these four letters, where two are identical, is done in $\frac{4!}{2!} = 12$ ways. Combining these operations, the total number of words is $2 \times 10 \times 12 = 240$.

The relationship between binomial coefficients can be analyzed by simplifying the ratio of successive terms to isolate the variable expression. Using the identity $\frac{{}^{36}C_{r+1}}{{}^{35}C_{r}} = \frac{36}{r+1}$, the given equation $6 \times {}^{35}C_{r} = (k^2 - 3) \times {}^{36}C_{r+1}$ rearranges to $k^2 - 3 = \frac{r+1}{6}$. For $r=5$, $k^{2-}3=1$ resulting in $k=\pm 2$, and for $r=35$, $k^{2-}3=6$ resulting in $k=\pm 3$, providing a total of 4 ordered pairs.

The probability of a single die showing a 3 or 5 is $p = \frac{1}{3}$, making the probability of not showing these values $q = \frac{2}{3}$. For four dice thrown independently, the probability that at least two show a 3 or 5 is calculated by $P(X \ge 2) = 1 - P(X=0) - P(X=1)$, resulting in $1 - (\frac{2}{3})^4 - 4(\frac{1}{3})(\frac{2}{3})^3 = \frac{33}{81}$. Since the experiment is performed 27 times, the expected value is $27 \times \frac{33}{81} = 11$.