Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The combinatorial identity $\,{}^{n}C_r \cdot {}^{n-r}C_{k-r} = {}^{n}C_k \cdot {}^{k}C_r$ simplifies the given summation by allowing us to rewrite each term $\,{}^{50}C_r \cdot {}^{50-r}C_{25-r}$ as $\,{}^{50}C_{25} \cdot {}^{25}C_r$. Factoring the constant term $\,{}^{50}C_{25}$ outside the summation results in $\,{}^{50}C_{25} \sum_{r=0}^{25} \,{}^{25}C_r$, where the sum of the combinations equals $2^{25}$, establishing that K is $2^{25}$.

Each pair of the $m$ men plays two games, resulting in $2 \times \binom{m}{2} = m(m-1)$ total games among men, while the games between $m$ men and 2 women total $m \times 2 \times 2 = 4m$. Setting the difference to 84 gives $m^2 - m - 4m = 84$, which simplifies to $m^2 - 5m - 84 = 0$. Solving this quadratic equation yields $m=12$ as the positive solution.

The total number of ways to form a team of 2 girls and 3 boys from 5 girls and 7 boys is calculated by $\,{}^{5}C_2 \cdot {}^{7}C_3$. To exclude cases where specific boys A and B are together, we subtract the forbidden teams where both are included, which is calculated as $\,{}^{5}C_2 \cdot {}^{5}C_1$, representing the selection of 2 girls and the remaining required boy from the other 5. Subtracting these values, $350 - 50$ yields 300 possible teams.

For any set of three distinct labels chosen from the ten available, there is exactly one way to arrange them in increasing order such that $n_1 < n_2 < n_3$. This means the problem is equivalent to choosing any three balls from ten, calculated by the combination $\binom{10}{3} = \frac{10 \times 9 \times 8}{6} = 120$.

Divisibility by 11 requires that the alternating sum of the digits in a 6-digit number be 0 or a multiple of 11. For the set $\{0, 1, 2, 5, 7, 9\}$, the sum of all digits is 24, implying that the sum of digits in the odd positions must equal the sum of digits in the even positions, both being 12. Using the sets $\{9, 2, 1\}$ and $\{7, 5, 0\}$, there are 36 ways to arrange them when $\{7, 5, 0\}$ occupy even positions and 24 ways when they occupy odd positions (ensuring the leading digit is non-zero), resulting in a total of 60.

The area of a right-angled triangle with vertices at the origin, $(\alpha, 0)$, and $(0, \beta)$ is $\frac{1}{2} |\alpha \beta| = 50$, implying $|\alpha \beta| = 100$. Since $\alpha$ and $\beta$ must be integers, we identify that 100 has 9 divisors, and for each pair $(|\alpha|, |\beta|)$, there are 4 combinations of signs for the coordinates. This results in $9 \times 4 = 36$ total triangles.

The word BARRACK consists of 7 letters with frequency distribution A:2, R:2, B:1, C:1, K:1. Forming 4-letter words requires considering three mutually exclusive cases: all 4 letters are distinct, 2 letters are identical with 2 distinct, and 2 pairs of letters are identical. Calculating the arrangements for each case gives $120$, $144$, and $6$ permutations respectively, which sum to a total of 270.

The condition for a number to be divisible by $3$ is that the sum of its digits must be a multiple of $3$. Using the digits $0, 1, 2, 3, 4$ to form $4$-digit numbers between $2,000$ and $5,000$ requires the first digit to be $2, 3,$ or $4$. The sets of $4$ digits that sum to a multiple of $3$ are $\{0, 1, 2, 3\}$ and $\{0, 2, 3, 4\}$. For the set $\{0, 1, 2, 3\}$, starting with $2$ gives $3! = 6$ permutations and starting with $3$ gives $3! = 6$ permutations, totaling $12$ arrangements. For the set $\{0, 2, 3, 4\}$, starting with $2, 3,$ or $4$ each allows for $3! = 6$ permutations, contributing $18$ arrangements. Combining these cases results in $30$ possible numbers.

The total number of arrangements is determined by calculating the number of ways to select the items and then multiplying by the permutations possible under the given positional constraint. Selecting $4$ novels from $6$ and $1$ dictionary from $3$ provides $\binom{6}{4} \times \binom{3}{1} = 15 \times 3 = 45$ combinations. Placing the $1$ dictionary in the fixed middle position of the $5$ available spots leaves $4$ spots for the novels, which can be filled in $4! = 24$ ways, resulting in $45 \times 24 = 1080$ total arrangements.

The number of distinct $n$-digit numbers formed using only $3$ digits is given by $3^n$ according to the fundamental counting principle. We seek the smallest integer $n$ such that $3^n \ge 900$. Evaluating the powers of $3$, we find that $3^6 = 729$ and $3^7 = 2187$. Since $729$ is less than $900$ and $2187$ exceeds it, $7$ is the smallest integer that satisfies the requirement.

Arranging words in dictionary order requires identifying the sequence of letter permutations alphabetically. With the letters $E, E, N, Q, U$, words starting with $E$ yield $\frac{4!}{1!} = 24$ permutations. Words starting with $N$ provide $\frac{4!}{2!} = 12$ permutations. Moving to $Q$, words beginning with $QE$ contribute $3! = 6$ arrangements, and those starting with $QN$ contribute $\frac{3!}{2!} = 3$ arrangements. The word $QUEEN$ is the subsequent entry, making the total position $24 + 12 + 6 + 3 + 1 = 46$.

In a circular arrangement of $8$ people, the total number of ways to seat them is $(8-1)! = 7!$. To find the number of arrangements where two specific people are never adjacent, one subtracts the configurations where they sit together from the total. Treating the boy and girl as a single unit creates $7$ entities, which can be arranged in $(7-1)! = 6!$ ways, and the pair can be arranged internally in $2!$ ways, resulting in $2 \times 6!$ configurations where they are adjacent. Subtracting this from the total gives $7! - 2 \times 6! = 7 \times 6! - 2 \times 6! = 5 \times 6!$ ways.

The selection of 3 ladies and 3 men from the friends of X and Y is partitioned by the number of ladies chosen from X's friends, $k$. By selecting $k$ ladies and $3-k$ men from X's 4 ladies and 3 men, we must select the remaining $3-k$ ladies and $k$ men from Y's 3 ladies and 4 men. Summing these independent cases gives the total number of ways: ${}^4C_0 imes {}^3C_3 imes {}^3C_3 imes {}^4C_0 + {}^4C_1 imes {}^3C_2 imes {}^3C_2 imes {}^4C_1 + {}^4C_2 imes {}^3C_1 imes {}^3C_1 imes {}^4C_2 + {}^4C_3 imes {}^3C_0 imes {}^3C_0 imes {}^4C_3$. Calculating each term results in $1 + 144 + 324 + 16 = 485$.

To evaluate the sum $\sum_{r=1}^{10}(r^{2+}1)r!$, we rewrite the general term as $((r+1)^{2-}2r)r!$, which allows us to express it as $(r+1)(r+1)! - 2r \cdot r!$. By separating this into $\sum_{r=1}^{10} ((r+1)(r+1)! - r \cdot r!) - \sum_{r=1}^{10} r \cdot r!$, we utilize the telescoping sum property where $\sum_{r=1}^{10} r \cdot r! = \sum_{r=1}^{10} ((r+1)! - r!) = 11! - 1!$. Evaluating the entire expression yields $11 \cdot 11! - 1 - (11! - 1) = 10 \cdot 11!$.

Using the combination identity $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$, we simplify the general term for $n=15$ to $r^2 \cdot \frac{16-r}{r} = 16r - r^2$. We then compute the sum $\sum_{r=1}^{15} (16r - r^2)$ as $16 \sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2$. Substituting the standard summation formulas gives $16 \left(\frac{15 \times 16}{2}\right) - \left(\frac{15 \times 16 \times 31}{6}\right) = 1920 - 1240 = 680$.

The position of a word in a dictionary is determined by finding the number of words that precede it alphabetically, accounting for duplicate letters by dividing the total permutations by their factorials. By systematically calculating the arrangements of the letters S, M, A, L, and L starting with A, L, and M, we determine there are 48 words preceding those starting with S, and then by summing the permutations of words starting with SA and SL, we arrive at the precise rank of the word SMALL. This approach ensures all alphabetical possibilities are exhausted before reaching the specific target word.

To determine the total number of words of length four with fixed letters, we consider the arrangements of the remaining letters from the word "MEDITERRANEAN". Accounting for the multiplicities of the repeated letters that influence the permutations, the number of distinct arrangements of the available letters is given by the formula $\frac{11!}{(2!)^3}$. This formula correctly adjusts the total permutations for the frequency of the repeated letters within the multi-set.

We express the combination and permutation terms using factorials to simplify the given equation $\frac{{}^{n+2}C_6}{{}^{n-2}P_2} = 11$, resulting in $\frac{(n+2)!}{6!(n-4)!} \times \frac{(n-4)!}{(n-2)!} = 11$. This reduces to $\frac{(n+2)(n+1)n(n-1)}{720} = 11$, which simplifies to $(n+2)(n+1)n(n-1) = 7920$. Testing integer values for $n$ reveals that $n=9$ satisfies the equation, which also satisfies the quadratic equation $n^{2+}3n-108=0$.

To find the number of integers greater than 6000 using the digits 3, 5, 6, 7, and 8 without repetition, we consider both five-digit and four-digit cases. There are $5!$ possible five-digit integers, and for four-digit integers, the first digit must be 6, 7, or 8, meaning we have 3 choices for the first position and permute the remaining digits in $P(4,3)$ ways. Summing the 120 five-digit numbers and the 72 four-digit numbers yields the final total.

The Cartesian product $A \times B$ contains $2 \times 4 = 8$ elements, meaning there are $2^8$ total possible subsets. To find the number of subsets with 3 or more elements, we subtract the subsets with 0, 1, or 2 elements from the total, using the combination formula $\binom{n}{r}$ to find $\binom{8}{0} + \binom{8}{1} + \binom{8}{2}$. Subtracting the sum of these combinations, 1 + 8 + 28, from 256 leaves the required number of subsets.