Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

This arrangement problem is solved using the stars and bars method by treating the 11 blue cubes as dividers and the 5 red cubes as items to be placed. With 4 required gaps between the red cubes, each must contain at least 2 blue cubes, consuming 8 of the 11 blue cubes. The remaining 3 blue cubes are then distributed across the 6 available spaces, which is calculated as $\binom{3+6-1}{6-1}$, resulting in 56 ways.

Three-digit numbers with exactly one digit repeated twice are categorized by whether the repeated digit is zero or non-zero. When non-zero digits are involved, there are 9 choices for the repeated digit, 8 choices for the distinct digit, and 3 positional arrangements for each set, totaling 216. Including the cases where zero is either the repeated digit or the single distinct digit adds 27 further possibilities, leading to a sum of 243.

This distribution problem is modeled using generating functions, where we seek the coefficient of $x^{30}$ in the product of the polynomials representing the constraints for each child. Each child's allotment corresponds to a geometric series, and by setting up the product $(x^{4+}x^{5+}x^{6+}x^7)(x^{2+}x^{3+}x^{4+}x^{5+}x^6)(1+x+x^{2+}\dots)^2$, the constraint on $x^{30}$ is extracted. Evaluating the resulting coefficient using the binomial theorem yields 430 ways to distribute the candies.

For a $3 \times 3$ matrix with entries from $\{-1, 0, 1\}$ having a total sum of 5, we identify the specific counts of each entry. Representing the number of $1$s, $-1$s, and $0$s as $x, y,$ and $z$ respectively, the conditions $x+y+z=9$ and $x-y=5$ allow for three distinct distributions of entries. Calculating the permutations for these cases produces 36, 252, and 126 arrangements, respectively, which sum to 414.

Divisibility by 4 depends on the last two digits of a number, and given the range 1000 to 3000, the first digit must be either 1 or 2. When the first digit is 1, the digits {2, 3, 4, 5, 6} allow for 6 possible pairs at the end (24, 32, 36, 52, 56, 64), and each pair leaves 3 options for the second digit, resulting in $6 imes 3 = 18$ numbers. Similarly, when the first digit is 2, the available digits {1, 3, 4, 5, 6} permit 4 valid pairs (16, 36, 56, 64), which gives $4 imes 3 = 12$ numbers. Combining these two cases yields a total of 30 valid numbers.

The problem is solved using the Principle of Inclusion-Exclusion to count permutations containing a sequence of three consecutive integers. By defining sets $A$ and $B$ where the first three or last three elements are consecutive integers, the total count is found by calculating $n(A) + n(B) - n(A \cap B)$. This systematically accounts for all arrangements of the four chosen elements while excluding the overlapping cases to reach the total of 18915.

The total number of 3-digit numbers whose greatest common divisor with 36 is 2 is determined by considering numbers of the form $2k$, where $k$ must be an integer such that the greatest common divisor of $k$ and 18 is 1. For 3-digit numbers ranging from 100 to 999, the value of $k$ must lie in the interval $[50, 499]$. Calculating the number of such integers $k$ by excluding multiples of 2 and 3 within this range yields 150.

A number is divisible by 15 if it is a multiple of 5 and 3, which requires the last digit to be 5 and the sum of all digits to be divisible by 3. The sum of the available digits {1, 2, 3, 4, 5, 6, 7} is 28, so forming a 6-digit number requires excluding one digit $x$ such that $28 - x$ is a multiple of 3, meaning $x$ can be 1, 4, or 7. For each of these three scenarios, the remaining 5 digits are arranged in the first five positions, each contributing $5! = 120$ arrangements, which yields a total of 360 valid numbers.

A number is divisible by 11 if the absolute difference between the sums of its digits at odd and even positions is a multiple of 11. With the digit set {1, 2, 3, 4, 5, 7, 9} having a total sum of 31, we solve for $S_1 - S_2 = 11$ or $S_1 - S_2 = -11$ to find pairs of sums $(21, 10)$ or $(10, 21)$. Calculating the permutations for these cases yields 432 arrangements for the first case and 144 for the second, resulting in a total of 576.

To identify 3-digit odd numbers whose sum of digits is a multiple of 7, we set the units digit to $1, 3, 5, 7,$ or $9$ and determine the combinations for the hundreds and tens digits that satisfy the divisibility condition. For each possible unit digit, we find the number of valid pairs for the remaining positions based on the sum requirement. Summing these counts for each case yields 12, 13, 14, 12, and 12, which results in a total of 63.

The total number of such matrices is determined by the number of entries equal to 1, which must result in a prime sum. With 9 total slots, if k entries are 1, the sum is k, where k must be a prime number from the set {2, 3, 5, 7}. Calculating the combinations for each prime, we find the total number of matrices is given by the sum $\binom{9}{2} + \binom{9}{3} + \binom{9}{5} + \binom{9}{7} = 36 + 84 + 126 + 36 = 282$.

A number is divisible by 55 if it is divisible by both 5 and 11. Given the digits {2, 3, 4, 5, 6} without repetition, divisibility by 5 requires the last digit to be 5, and for a four-digit number $abcd$ with $d=5$, the condition for divisibility by 11 requires $(a+c) - (b+5) = 11k$. Testing the digits, we find the only valid condition is $a+c = b+5$, which yields 6 possible combinations-(3,4,2), (4,3,2), (2,6,3), (6,2,3), (3,6,4), and (6,3,4)-while no 5-digit arrangements of these specific digits satisfy the requirements.

To determine the number of subsets whose sum is not a multiple of 3, we first group the elements of $S$ by their remainder when divided by 3: $R_0 = \{3, 6, 9\}$, $R_1 = \{1, 4\}$, and $R_2 = \{2, 5\}$. A subset's sum is a multiple of 3 if the sum of the remainders of its elements is equivalent to 0 modulo 3. By calculating the number of possible subsets of each size $k$ from 1 to 6 that are not divisible by 3, we find the frequencies to be 4, 14, 22, 22, 14, and 4 respectively. Summing these individual counts yields a total of 80.

The prime factorization of the given number is $2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13}$. A divisor of the form $4n+1$ must be odd, which implies the exponent of 2 must be 0. Considering the prime factors $5$, $11$, and $13$, we note that $5 \equiv 1 \pmod{4}$ and $13 \equiv 1 \pmod{4}$, so any power of 5 or 13 will result in a value equivalent to 1 modulo 4. However, $11 \equiv -1 \pmod{4}$, so $11^c$ is equivalent to $1$ modulo 4 only when the exponent $c$ is even. There are 11 choices for the exponent of 5 ($0$ to $10$), 6 choices for the even exponent of 11 ($0, 2, 4, 6, 8, 10$), and 14 choices for the exponent of 13 ($0$ to $13$), resulting in $11 \times 6 \times 14 = 924$ valid divisors.

The central number in each figure is determined by calculating the difference between the two outer numbers and raising that difference to the power of its own factorial. For instance, when the outer numbers are 7 and 4, the difference is 3, which leads to the calculation $3^{3!} = 3^6$. Applying this same logic to the final set of 5 and 3, the difference is 2, which results in $2^{2!} = 2^2$, ultimately giving a final value of 4.

Triangles are formed by selecting one point from each of three different sides of the rectangle, ensuring the points are not collinear. Calculating the combinations for all four possible sets of three sides-$5 \times 7 \times 6$, $5 \times 7 \times 9$, $5 \times 6 \times 9$, and $7 \times 6 \times 9$-gives $\alpha = 210 + 315 + 270 + 378 = 1173$. Quadrilaterals are formed by selecting one point from each of the four sides, calculated as $\beta = 5 \times 7 \times 6 \times 9 = 1890$. The difference is $1890 - 1173 = 717$.

Dividing 10 students into three groups A, B, and C where each group is non-empty and C has at most three members requires selecting the members for C first, then distributing the remaining students between A and B. For a fixed size $k$ of group C, the number of ways is $\binom{10}{k}(2^{10-k} - 2)$, accounting for the constraint that A and B must both have at least one student. Summing this for $k=1, 2, 3$ yields $10(512-2) + 45(256-2) + 120(128-2) = 5100 + 11430 + 15120 = 31650.$

To select a team of 11 from 15 players consisting of 6 bowlers, 7 batsmen, and 2 wicketkeepers under the constraints of at least 4 bowlers, 5 batsmen, and 1 wicketkeeper, we analyze the possible distributions $(B, Ba, W)$. The valid combinations summing to 11 are $(5, 5, 1)$, $(4, 6, 1)$, and $(4, 5, 2)$. Calculating the number of ways for each using combinations yields ${}^{6}C_{5} \cdot {}^{7}C_{5} \cdot {}^{2}C_{1} = 252$, ${}^{6}C_{4} \cdot {}^{7}C_{6} \cdot {}^{2}C_{1} = 210$, and ${}^{6}C_{4} \cdot {}^{7}C_{5} \cdot {}^{2}C_{2} = 315$. Adding these values together gives a total of $777$ ways.

The number of odd divisors of an integer $n = 2^x 3^y 5^z$ depends only on the exponents of the odd prime factors, calculated as $(y+1)(z+1)$. Given $y+z=5$ and the reciprocal sum $y^{-1} + z^{-1} = 5/6$, the equations simplify to $y+z=5$ and $yz=6$, which implies $y=3$ and $z=2$ since $y>z$. Applying these exponents to the divisor formula results in $(3+1)(2+1) = 4 \times 3 = 12$.

Using the definitions ${}^{n}P_{r} = \frac{n!}{(n-r)!}$ and ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$, we start by equating ${}^{n}P_{r}={}^{n}P_{r+1}$, which simplifies to $\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}$, leading to $n-r = 1$. Next, we use the identity ${}^{n}C_{r}={}^{n}C_{r-1}$, which expands to $\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}$, simplifying to $n-r+1 = r$ or $n+1 = 2r$. Substituting $n = r+1$ into the second equation yields $r+2 = 2r$, which results in $r=2$.