Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The number of integers divisible by a set of divisors is found using the principle of inclusion-exclusion applied to the total count of three-digit numbers. The set of three-digit numbers divisible by 6 (the least common multiple of 2 and 3) is given by rac{900}{6} = 150, while the set divisible by 36 (the least common multiple of 4 and 9) is given by rac{900}{36} = 25. Since every number divisible by 36 is necessarily divisible by 6, the number of integers divisible by 2 and 3 but not 4 and 9 is the difference between these two counts, resulting \in $150 - 25 = 125$.

Let $m=6a$ and $n=6b$ with $a < b$, where $m$ and $n$ are $2$-digit numbers, constraining $a$ and $b$ such that $2 ext{ } ext{le} ext{ } a < b ext{ } ext{le} ext{ } 16$. The condition $ext{gcd}(m, n)=6$ is equivalent to requiring $ext{gcd}(a, b)=1$. By counting the number of integers $b$ coprime to each $a$ in the range $2 ext{ } ext{le} ext{ } a < b ext{ } ext{le} ext{ } 16$, we find the total number of such pairs is $64$.

Dictionary order relies on alphabetic sorting of the available letters $A, K, N, P, R, U$. We determine the position by calculating permutations starting with each character until we accumulate $440$ words. Starting with $A, K,$ or $N$ accounts for $3 imes 120 = 360$ words. Within the $P$-section, words starting with $PA, PK,$ or $PN$ contribute $3 imes 24 = 72$ additional words, reaching $432$. We then look for the $8^{ ext{th}}$ word in the $PR$ sequence, where the remaining letters are $A, K, N, U$. Counting $PR A...$ provides $6$ words, ending at $438$, then $PR K A N U$ is $439$, and finally $PR K A U N$ is the $440^{ ext{th}}$ entry.

Partitioning a set of 9 elements into 3 subsets of 3 elements each involves the multinomial coefficient $\frac{9!}{3!3!3!}$. Since the question asks for the total number of partitions for the set $S$, computing this value provides the total distribution of elements into the specified groups. This calculation results in 1680 possible partitions.

The number of words formed from a set of letters with repetitions is found by categorizing selections based on frequency patterns, such as selecting 3 identical letters with 1 distinct or 2 identical with 2 distinct. For the letters of 'DISTRIBUTION', these cases result in permutations of $32, 6, 672,$ and $3024$ respectively. Summing these permutations yields a total of 3734.

The number of non-negative integer solutions to a linear equation is found by systematically evaluating the equation for each possible value of one variable. For the equation $x+2y+3z=42$, we fix $z$ as an integer from 0 to 14, and for each $z$, the number of non-negative solutions for $x$ and $y$ is $\lfloor \frac{42-3z}{2} \rfloor + 1$. Summing these values for every valid $z$ leads to the total of 169.

Combinatorial selection under constraints is determined by partitioning the total requirement into all valid distributions across sections A, B, and C that satisfy the minimum quota of 4. Summing the products of combinations for each valid partition, such as $\binom{8}{7}\binom{6}{4}\binom{6}{4}$ and $\binom{8}{5}\binom{6}{5}\binom{6}{5}$, accounts for every possible selection. Performing these calculations across all cases leads to 11376.

Let the number be represented by digits $a, b, c$ such that $a+b+c=14$, where $1 ot ext{le} a ot ext{le} 9$ and $0 ot ext{le} b, c ot ext{le} 9$. The total non-negative solutions to $a-1+b+c=13$ is given by inom{13+3-1}{3-1} = inom{15}{2} = 105. We must subtract cases violating the digit constraints: $a ot ext{ge} 10$ yields inom{4+2}{2} = 15 cases, while $b ot ext{ge} 10$ and $c ot ext{ge} 10$ each yield inom{3+2}{2} = 10 cases. The final count is $105 - (15 + 10 + 10) = 70$.

Generating functions represent the count of possible outcomes for problems involving the sum of variables across multiple independent trials, where the number of ways to achieve a sum is the coefficient of the corresponding power in an expansion. Here, we find the coefficient of $x^{16}$ in the expansion of $(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^4$, which is equivalent to $x^4(1 - x^6)^4(1 - x)^{-4}$. Expanding this expression, we seek the coefficient of $x^{12}$ in the product (1 - 4x^6 + 6x^{12})(1 + 4x + 10x^2 + inom{n+3}{3}x^n). Calculating the specific term $1 \cdot \binom{12+3}{3} - 4 \cdot \binom{6+3}{3} + 6 \cdot \binom{0+3}{3}$ yields $455 - 336 + 6 = 125$.

The number of distinct permutations for a multiset is calculated by dividing the factorial of the total number of items by the factorial of each repeating element count. For the letters B, B, H, J, O, there are $5!/2! = 60$ total permutations. Alphabetical order starts with B, H, J, O. There are $4!/1! = 24$ words starting with B, $4!/2! = 12$ words starting with H, and $4!/2! = 12$ words starting with J, totaling 48 words. The 49th and 50th words start with OB, and by arranging the remaining letters B, H, and J, we find the 49th word is OBBHJ and the 50th word is OBBJH.

Arranging the letters of NAGPUR in alphabetical order as A, G, N, P, R, U allows us to determine the position of a specific word by counting all preceding permutations. By systematically calculating the number of words starting with A, G, NA, NG, NP, and finally NRA, we reach the total before arriving at the target word. Following this lexicographical counting, the 315th word in the sequence is identified as NRAPGU.

The word MATHEMATICS has $11$ letters with frequencies: $M=2, A=2, T=2, H=1, I=1, C=1, S=1, E=1$. We need to choose $5$ letters, classified by the number of pairs: choosing $5$ distinct letters from the $8$ unique options gives inom{8}{5} = 56 ways. Choosing $1$ pair from the $3$ available pairs and $3$ other distinct letters yields 3 imes inom{7}{3} = 3 imes 35 = 105 ways. Choosing $2$ pairs from the $3$ available and $1$ other distinct letter yields 3 imes inom{6}{1} = 18 ways, leading to a total of $56 + 105 + 18 = 179$.

Distributing distinct objects into indistinguishable containers corresponds to the partition of a set into subsets, which is counted using Stirling numbers of the second kind, denoted $S(n, k)$. Since the offices are indistinguishable and can remain empty, we sum the ways to partition 5 employees into 1, 2, 3, or 4 non-empty subsets. This is calculated as $S(5, 1) + S(5, 2) + S(5, 3) + S(5, 4)$, which evaluates to $1 + 15 + 25 + 10 = 51$.

Partitioning 8 identical books into 4 identical shelves corresponds to finding the number of partitions of 8 into at most 4 parts. By systematically listing the non-increasing sequences $(a, b, c, d)$ such that $a+b+c+d=8$ with $a eq b eq c eq d eq 0$ not strictly required, but considering $a eq b eq c eq d eq 0$ is not necessary, we account for all possible shelf arrangements. Categorizing by the number of empty shelves, the sum of these exhaustive cases results in 15 distinct arrangements.

The relationship between binomial coefficients is defined by the ratio formula $\frac{{}^{n}C_{r}}{{}^{n-1}C_{r-1}} = \frac{n}{r}$. Using the given ratios $\frac{{}^{n+1}C_{r+1}}{{}^{n}C_{r}} = \frac{11}{7}$ and $\frac{{}^{n}C_{r}}{{}^{n-1}C_{r-1}} = \frac{5}{3}$, we derive the linear equations $7n = 11r+4$ and $3n = 5r$. Solving this system yields $n=10$ and $r=6$, which results in $2n+5r = 2(10) + 5(6) = 50$.

Lexicographical ordering determines the rank of a word by counting all permutations that precede it alphabetically. For the word "GTWENTY" using the letters $\{E, G, N, T, T, W, Y\}$, we sum the permutations of words starting with E, GE, and GN, and then consider the sub-permutations for words starting with GT. Counting these systematically leads to the final rank of 553.

Geometry-based counting problems involving points on lines are solved by calculating the total possible combinations and subtracting the subsets of points that are collinear, as three points on a line cannot form a triangle. The total number of ways to choose any 3 points from 18 is $\binom{18}{3} = 816$. We then subtract the sets of 3 points that lie entirely on the sides AB, BC, and CA, which are $\binom{5}{3}$, $\binom{6}{3}$, and $\binom{7}{3}$ respectively. Performing the subtraction $816 - 10 - 20 - 35$ results in $751$.

A triangle formed from the vertices of a regular octagon has no common side if none of its edges coincides with the polygon's sides. By taking the total number of triangles $\binom{8}{3} = 56$ and subtracting those that share exactly one side ($8 \times 4 = 32$) or two sides ($8 \times 1 = 8$), we isolate the cases with no common sides. This calculation results in $56 - 40 = 16$ valid triangles.

Pascal's identity, ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$, simplifies the inequality to determine $m=2$, while the definition of permutations ${}^nP_r = \frac{n!}{(n-r)!}$ allows us to solve the ratio for $n=8$. Substituting these into the final expression ${}^8P_3 + {}^9C_2$ gives $8 \times 7 \times 6 + 36$, which equals 372.

Distributing identical items with a minimum quota per recipient is solved by first fulfilling the requirement and then using the stars and bars theorem. With 21 apples distributed to three children such that each receives at least two, we first allocate 6 apples, leaving 15 to be distributed as $\binom{15+3-1}{3-1} = \binom{17}{2}$. Calculating this combination results in 136.