Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The divisibility of a number by 3 is determined by the sum of its digits, which must be a multiple of 3. With the set of digits {2, 3, 4, 5, 7}, the total number of three-digit permutations is $P(5, 3) = 60$. By identifying the specific triplets that do not sum to a multiple of 3 and accounting for their distinct permutations, one finds that there are 36 such three-digit numbers.

The expression $\alpha = \frac{24!}{(4!)^6}$ represents $6!$ times the number of ways to partition 24 distinct objects into 6 unlabeled groups of 4 objects each. Because the number of ways to partition distinct objects must be an integer, $\alpha$ is clearly a natural number. A parallel argument applies to $\beta = \frac{120!}{(5!)^{24}}$, which relates to the partition of 120 objects into 24 groups of 5, confirming that both $\alpha$ and $\beta$ belong to $\mathbb{N}$.

Combinatorial selection problems involving multiple groups require splitting the objective into mutually exclusive cases based on the number of items drawn from each group. We define $m_1$ as the number of men selected from Group A, which necessitates selecting $4-m_1$ women from Group A, $4-m_1$ men from Group B, and $m_1$ women from Group B. Calculating the product of these combinations $\binom{4}{m_1}\binom{5}{4-m_1} \cdot \binom{5}{4-m_1}\binom{4}{m_1}$ for each valid value of $m_1$ from 0 to 4 and summing them gives $25 + 1600 + 3600 + 400 + 1 = 5626$.

The prime factors of $2310$ are $A = \{2, 3, 5, 7, 11\}$. Calculating $f(x) = \lfloor \log_2 (x^2 + \lfloor \frac{x^3}{5} \rfloor ) \rfloor$ for each: $f(2) = \lfloor \log_2 5 \rfloor = 2$, $f(3) = \lfloor \log_2 14 \rfloor = 3$, $f(5) = \lfloor \log_2 50 \rfloor = 5$, $f(7) = \lfloor \log_2 117 \rfloor = 6$, and $f(11) = \lfloor \log_2 387 \rfloor = 8$. The range of the function is $B = \{2, 3, 5, 6, 8\}$, which also has $5$ elements. The number of one-to-one functions from a set of $5$ elements to another set of $5$ elements is $5! = 120$.

To determine the number of permutations avoiding specific substrings, use the principle of inclusion-exclusion by subtracting the arrangements containing the forbidden strings from the total $7!$ permutations. The number of arrangements containing the string $153$ is $5!$, while those containing $2467$ result in $4!$ arrangements. Accounting for the overlap where both strings appear by subtracting $2!$, the final calculation is $5040 - (120 + 24 - 2)$, resulting in $4898$ valid permutations.

The number of non-negative integer solutions to $x+y+z=15$ is found using the stars and bars formula inom{n+r-1}{r-1}, which gives inom{15+3-1}{3-1} = inom{17}{2} = 136. To find the number of solutions with distinct integers, we subtract cases where variables are not distinct: cases where all three variables are equal, such as $(5, 5, 5)$, and cases where exactly two are equal. There is only $1$ case where all three are equal, and for the cases where exactly two are equal, we find $7$ possible values for the repeating pair $(x, x, z)$ excluding the $(5, 5, 5)$ case, which gives $21$ total permutations when considering the position of the distinct variable, leading to $136 - 21 - 1 = 114$ distinct solutions.

A five-digit number is divisible by 5 if it ends in 0 or 5 and is greater than 40000 if it starts with 5, 7, or 9. We analyze the possibilities based on the first and last digits, ensuring no repetition of the set {0, 1, 3, 5, 7, 9}. If the first digit is 5, the last must be 0, allowing $4 \times 3 \times 2 = 24$ arrangements; if the first is 7 or 9, the last can be 0 or 5, providing two scenarios for each, each yielding 24 arrangements. Summing these, $24 + 24 + 24 + 24 + 24$ gives the final total of 120.

The sum of all four-digit numbers formed by the digits $1, 2, 2, 3$ is determined by calculating the sum of the digits at each position. The total number of unique permutations is rac{4!}{2!} = 12, meaning each digit appears $3$ times at each position, with adjustments for the repeated digit $2$. The sum of digits at any given position is $1(3) + 2(6) + 3(3) = 24$, so the total sum is $24 imes (1000 + 100 + 10 + 1)$, which equals $24 imes 1111 = 26664$.

The number of arrangements where two letters do not appear together is found by subtracting the cases where they are adjacent from the total possible arrangements. For the word MATHEMATICS, the total permutations are $\frac{11!}{2!2!2!}$, and treating the pair C and S as a single unit results in $\frac{10!}{2!2!2!} \times 2!$ arrangements. Factoring out $6!$ from this difference yields $5670 \times 6!$, indicating that the required value $k$ is $5670$.

The increasing order of six-digit numbers with distinct digits from $1$ to $9$ is determined by combinations. Starting with $1$ gives ${}^{8}C_{5} = 56$ numbers, and starting with $2$ gives ${}^{7}C_{5} = 21$ numbers, reaching a cumulative total of $77$. Since the $72^{\text{th}}$ number starts with $2$, we note that the numbers starting with $23$ account for $15$ values, which places the $72^{\text{th}}$ number within the sequence starting with $24$. The smallest such value following the increasing rule is $245678$, and adding these digits yields $32$.

First, select 2 vowels from the available 3 in $\binom{3}{2} = 3$ ways and 2 consonants from the 4 available in $\binom{4}{2} = 6$ ways. These 4 distinct letters can then be arranged in $4! = 24$ different sequences. Multiplying these counts gives a total of $3 \times 6 \times 24 = 432$ words.

The word ASSASSINATION contains 13 letters, consisting of 6 vowels (A, A, A, I, I, O) and 7 consonants (S, S, S, S, N, N, T). To keep all vowels together, we treat the vowel group as a single entity, resulting in 8 total items to arrange, which can be done in $\frac{8!}{4!2!1!1!}$ ways, where the repetitions are four S's and two N's. Within the vowel group, the 6 letters can be arranged in $\frac{6!}{3!2!1!}$ ways, and multiplying these two counts provides the final total of $840 \times 60 = 50400$.

The permutation formula is ${}^nP_r = \frac{n!}{(n-r)!}$. Expanding the given ratio using this formula leads to $\frac{(2n+1)!}{(n+2)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$. By simplifying the factorials, we obtain $\frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{11}{21}$, which reduces to $\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$. Solving this quadratic equation yields $n=5$, and substituting this into $n^{2+}n+15$ gives $25+5+15=45$.

A three-digit number is divisible by 3 if the sum of its digits is a multiple of 3. Using the digits $\{1, 3, 5, 8\}$, we categorize combinations by their residues modulo 3, such as $(1,1,1)$, $(3,3,3)$, $(5,5,5)$, $(8,8,8)$, $(5,5,8)$, $(8,8,5)$, $(1,3,5)$, and $(1,3,8)$. Counting the distinct permutations for these sets leads to a total of 22 valid numbers.

The divisibility of $x+y$ by 5 depends on the remainders of $x$ and $y$ when divided by 5. We partition the integers 1 to 25 into five sets $R_k$ containing 5 elements each, corresponding to remainders $k=0,1,2,3,4$. The condition $x+y \equiv 0 \pmod 5$ holds when $x, y \in R_0$ (with $x \neq y$), or pairs $(R_1, R_4)$, $(R_4, R_1)$, $(R_2, R_3)$, and $(R_3, R_2)$. This yields $5 \times 4 = 20$ ways from $R_0$, and $4 \times (5 \times 5) = 100$ ways from the other pairs, totaling 120 ways.

To find the count of 3-digit numbers divisible by 3 or 4 but not 48, we use the principle of inclusion-exclusion. Total 3-digit numbers divisible by 3 is 300, by 4 is 225, and by both 12 is 75, giving 300 + 225 - 75 = 450 numbers divisible by either 3 or 4. We then subtract the numbers divisible by 48, which are 48 * k in the range [100, 999], specifically from k = 3 to k = 20, totaling 18 values. Subtracting these from the previous set gives 450 - 18 = 432.

Lattice points strictly inside the region bounded by the coordinate axes and $3x+4y=60$ must satisfy the inequality $3x+4y < 60$ for positive integers $x$ and $y$. Since $y$ is a multiple of $x$, we express $y=mx$ and substitute into the inequality to get $x(3+4m) < 60$. Evaluating this for each positive integer $m$ allows us to determine the possible values for $x$ in each case, and summing the counts for all $m$ yields a total of $31$ valid points.

The count of numbers is found using the inclusion-exclusion principle, which states $|A \cup B| = |A| + |B| - |A \cap B|$. The number of multiples of 3 in the range 1000 to 2800 is 600, while the multiples of 11 in that same range are found by subtracting those from 1 to 999 from the total multiples up to 2799, resulting in 164. Subtracting the intersection of multiples of 33, identified as 54 values, yields a total of $600 + 164 - 54 = 710$.

The selection of 5 courses from 12, where 5 are language courses and 7 are non-language, subject to choosing at most two language courses, uses the combination formula $\binom{n}{r}$. We sum the cases for exactly 0, 1, and 2 language courses: $\binom{5}{0} \binom{7}{5} + \binom{5}{1} \binom{7}{4} + \binom{5}{2} \binom{7}{3} = 21 + 175 + 350 = 546$.

A square matrix of order 5 with entries from $\{0,1\}$ having row and column sums of 1 is a permutation matrix. Placing a 1 in each row such that no two 1s share a column corresponds to creating a bijection from row indices to column indices. The number of such configurations is the number of permutations of 5 distinct elements, given by $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.