Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

Counting non-negative integer solutions to $a + b + 2c = 22$ is simplified by fixing the variable $c$ as an integer $k$ and applying the stars and bars logic to the remaining equation $a + b = 22 - 2k$. For a sum $S$, the number of non-negative solutions is $S + 1$, so for each $k$, we have $22 - 2k + 1 = 23 - 2k$ solutions. Summing this expression for all possible values of $k$ from $0$ up to $11$, we compute $\sum_{k=0}^{11} (23 - 2k) = 23 \times 12 - 2 \times \frac{11 \times 12}{2} = 276 - 132 = 144$.

The number of strictly increasing functions from a set of size $n$ to a set of size $m$ is given by the binomial coefficient inom{m}{n}. When the condition $f(i) eq i$ is imposed for all $i ext{ in the domain } 1, 2, ext{...}, n$, the count reduces to the binomial coefficient inom{m-1}{n}. Substituting $n=6$ and $m=9$, we calculate inom{8}{6}, which equals inom{8}{2} = 28.

With one batsman and one bowler already fixed as captain and vice-captain, we must select the remaining $8$ players from the $11$ available candidates to complete the team of $10$. To satisfy the requirement of at least $4$ batsmen and $4$ bowlers total, we need to pick $3$ to $5$ additional batsmen and $3$ to $5$ additional bowlers such that their total count is $8$. The number of ways to pick these is inom{6}{3}inom{5}{5} + inom{6}{4}inom{5}{4} + inom{6}{5}inom{5}{3}, which results in $20 + 75 + 60 = 155$ total ways.

In the set of all seven-digit numbers ranging from $1,000,000$ to $9,999,999$, the distribution of odd and even digit sums is perfectly uniform, meaning exactly half of these numbers satisfy the condition. The total count is $\frac{9 \times 10^6}{2} = 4.5 \times 10^6$, which is equivalent to $9 \times 5 \times 10^5$. Comparing this to the expression $m \cdot n \cdot 10^n$, we find $m=9$ and $n=5$, giving a sum of 14.

For a function $f: \{1, 2, \dots, 100\} \longrightarrow \{0, 1\}$, the requirement that exactly one element \in the subset $\{1, \dots, 98\}$ maps to 1 implies there are $\binom{98}{1} = 98$ ways to select which element maps to 1. The remaining elements \in that subset must map to 0, which is determined \in one way. The remaining two elements of the domain, 99 and 100, can each independently map to either 0 or 1, providing $2^2 = 4$ combinations of assignments. Multiplying these independent choices gives $98 \times 4 = 392$ total functions.

To select 4 boys and 4 girls with exactly 5 people from group A (7 boys, 3 girls) and 3 from group B (6 boys, 5 girls), we analyze the mutually exclusive scenarios based on the gender distribution from each group. Selecting 4 boys and 1 girl from A and 0 boys and 3 girls from B yields $\binom{7}{4} \binom{3}{1} \times \binom{6}{0} \binom{5}{3} = 1050$ ways. Selecting 3 boys and 2 girls from A and 1 boy and 2 girls from B yields $\binom{7}{3} \binom{3}{2} \times \binom{6}{1} \binom{5}{2} = 6300$ ways. Selecting 2 boys and 3 girls from A and 2 boys and 1 girl from B yields $\binom{7}{2} \binom{3}{3} \times \binom{6}{2} \binom{5}{1} = 1575$ ways. Summing these values gives a total of 8925 ways.

To construct sequences of ten terms containing exactly five 1s, three 2s, and two 0s, we calculate the number of unique arrangements for this multiset. The total number of ways is given by the multinomial coefficient $\frac{10!}{5!3!2!}$, which evaluates to $2520$.

We classify the $5$-digit numbers starting with $a \in \{5, 6, 7\}$ by checking the condition $a+e \le 8$ for the last digit $e$. For $a=5$, $e \in \{0, 1, 2, 3\}$ allows $4 \times 8^3$ numbers, for $a=6$, $e \in \{0, 1, 2\}$ allows $3 \times 8^3$ numbers, and for $a=7$, $e \in \{0, 1\}$ allows $2 \times 8^3$ numbers. Summing these possibilities gives $(4+3+2) \times 8^3 = 4608$ configurations. Since the problem requires numbers strictly greater than $50000$, we exclude the value $50000$ where $a=5, b=c=d=e=0$ results in $5+0=5 \le 8$, leaving $4607$ valid numbers.

The problem requires distributing five distinct letters into eight boxes spread across three rows, with the constraint that every row must contain at least one letter and no box can hold more than one. This implies we should first determine the number of ways to select five distinct boxes from the total eight, ensuring that the selection satisfies the row-occupancy condition, and then account for the distinct nature of the letters by arranging them within those chosen boxes.

The number of boxes occupied in the first, second, and third rows can be represented by variables $x$, $y$, and $z$, such that $x+y+z=5$. Given the row sizes of three, three, and two, valid distributions are found by satisfying $1 \le x \le 3$, $1 \le y \le 3$, and $1 \le z \le 2$. The possible combinations for $(x, y, z)$ are $(3, 1, 1)$, $(2, 2, 1)$, $(1, 3, 1)$, $(2, 1, 2)$, and $(1, 2, 2)$. For each combination, the number of ways to choose the specific boxes is calculated by the product of combinations for each row, given by $\binom{3}{x} \binom{3}{y} \binom{2}{z}$.

Calculating these specific choices results in $\binom{3}{3} \binom{3}{1} \binom{2}{1} = 6$ for the first case, $\binom{3}{2} \binom{3}{2} \binom{2}{1} = 18$ for the second, $\binom{3}{1} \binom{3}{3} \binom{2}{1} = 6$ for the third, $\binom{3}{2} \binom{3}{1} \binom{2}{2} = 9$ for the fourth, and $\binom{3}{1} \binom{3}{2} \binom{2}{2} = 9$ for the final case. Summing these possibilities yields 48 unique ways to select the boxes. Since we are placing five distinct letters into these five selected boxes, we multiply the number of box combinations by the number of permutations of the letters, given by $5! = 120$, resulting in 5760 total arrangements.

To form a $7$-digit number with a digit sum of $11$ using only $1, 2,$ and $3$, we establish the constraints $n_1 + n_2 + n_3 = 7$ and $n_1 + 2n_2 + 3n_3 = 11$, where $n_i$ represents the count of digit $i$. Solving these simultaneous equations leads to three distinct digit sets: $\{1,1,1,1,2,2,3\}$ yielding $\frac{7!}{4!2!} = 105$ permutations, $\{1,1,1,2,2,2,2\}$ yielding $\frac{7!}{4!3!} = 35$ permutations, and $\{1,1,1,1,1,3,3\}$ yielding $\frac{7!}{5!2!} = 21$ permutations. Combining these mutually exclusive cases results in a total of $161$ possible numbers.

The total number of triangles formed by $n$ points where no three are collinear is calculated using the combination inom{n}{3}. When $k$ points are collinear, they cannot form a triangle, so we must subtract the non-existent triangles represented by inom{k}{3} from the total. Applying this formula for $12$ points with $5$ collinear, we calculate inom{12}{3} - inom{5}{3} = 220 - 10 = 210.

To arrange five chosen letters in alphabetical order with 'M' fixed as the middle character, we select two letters from the 12 available before 'M' and two from the 13 available after 'M'. Since the alphabetical order is inherent, the number of ways is the product of the independent combinations $\binom{12}{2} \times \binom{13}{2}$, resulting in $66 \times 78 = 5148$.

The values for n and r are determined using the ratio property of combinations, specifically rac{{}^nC_{r-1}}{{}^nC_r} = rac{r}{n-r+1} = rac{28}{56} = rac{1}{2} and rac{{}^nC_r}{{}^nC_{r+1}} = rac{r+1}{n-r} = rac{56}{70} = rac{4}{5}. Solving these simultaneous equations identifies $n=8$ and $r=3$, which sets the coordinates of vertex C to $(3(3)-8, 3^{2-}8-1) = (1, 0)$. The centroid $(x, y)$ of the triangle formed by vertices $A(4 ext{cos } t, 4 ext{sin } t)$, $B(2 ext{sin } t, -2 ext{cos } t)$, and $C(1, 0)$ is calculated as x = rac{4 ext{cos } t + 2 ext{sin } t + 1}{3} and y = rac{4 ext{sin } t - 2 ext{cos } t}{3}. Rearranging these equations to isolate the trigonometric terms and squaring both sides gives $(3x-1)^2 + (3y)^2 = (4 ext{cos } t + 2 ext{sin } t)^2 + (4 ext{sin } t - 2 ext{cos } t)^2 = 20( ext{cos}^2 t + ext{sin}^2 t) = 20$, confirming that $ext{\alpha } = 20$.

The problem is solved by calculating the arrangements for two distinct scenarios and summing the results. When all 5 boys sit together, they function as a single unit, which combined with 4 girls creates 5 entities that can be arranged in $5!$ ways, multiplied by $5!$ ways to arrange the boys internally, totaling $120 \times 120 = 14400$. When no two boys sit together, the 4 girls are arranged \in $4!$ ways, creating 5 gaps (including the ends) where the 5 boys can be placed \in $P(5, 5) = 5!$ ways, resulting \in $24 \times 120 = 2880$. Adding these two cases yields $14400 + 2880 = 17280$ total ways.

Given the probability distribution follows a geometric progression $P(W_n) = 2^{n-1}p$, and the sum of all probabilities must equal 1, we determine $p = \frac{1}{2^{120}-1}$. The rank of the word $CDBEA$ is found by summing permutations starting with A, B, CA, CB, CDA, and CDBA, which places it at the 64th position. Substituting $n=64$ into the probability formula yields $P(W_{64}) = \frac{2^{63}}{2^{120}-1}$, identifying $\alpha=63$ and $\beta=120$, leading to a sum of 183.

To find the number of integers in the range $(212, 999)$ with a digit sum of $15$, we represent the numbers as $d_1 d_2 d_3$ such that $d_1 + d_2 + d_3 = 15$. We categorize these by the hundreds digit $d_1 \in \{2, 3, \dots, 9\}$ and determine the number of valid pairs $(d_2, d_3)$ for each $d_1$ by solving $d_2 + d_3 = 15 - d_1$. Summing the counts for each case yields $6 + 7 + 8 + 9 + 10 + 9 + 8 + 7$, totaling $64$ numbers.

The problem requires choosing subsets of letters from $\{M, A, T, H, S\}$ such that every selected letter appears at least twice in a $6$-letter arrangement. We partition the solution based on the distribution of letter frequencies. Selecting three distinct letters to appear twice each contributes ${}^5C_3 \times \frac{6!}{2!2!2!} = 900$ ways. Choosing two letters, where one appears four times and the other twice contributes $10 \times 15 \times 2 = 300$, while choosing two letters appearing three times each contributes $10 \times 20 = 200$. Finally, using a single letter repeated six times adds $5$ possibilities, totaling $1405$ words.

Forming a triangle with 22 points requires selecting three vertices that are not collinear. Since points on line $L_1$ and line $L_2$ are collinear, triangles are formed by choosing either two points from $L_1$ and one from $L_2$, one point from $L_1$ and two from $L_2$, or one point from each line combined with the origin. This calculation results in $\binom{12}{2}\binom{9}{1} + \binom{12}{1}\binom{9}{2} + \binom{12}{1}\binom{9}{1}\binom{1}{1}$, which simplifies to $594 + 432 + 108 = 1134$.

To determine the number of arrangements where vowels are separated, the most efficient approach is to subtract the arrangements where vowels are together from the total permutations of the word. The word DAUGHTER contains $8$ distinct letters, including $3$ vowels and $5$ consonants, giving a total of $8! = 40320$ permutations. When all $3$ vowels are grouped together as a single entity, we arrange $6$ entities ($1$ vowel-block and $5$ consonants) in $6!$ ways, with the vowels themselves arranged internally in $3!$ ways, resulting in $720 imes 6 = 4320$ arrangements. Subtracting this from the total yields $40320 - 4320 = 36000$.

We treat the group of $3$ girls and $4$ boys as two distinct blocks, which can be arranged in $2! = 2$ ways (Girls-Boys or Boys-Girls). The girls can be arranged within their block in $3! = 6$ ways. For the boys, total arrangements are $4! = 24$, but we must exclude cases where $B_1$ and $B_2$ are adjacent; if we treat $(B_1B_2)$ as one unit, there are $3! imes 2! = 12$ invalid arrangements, leaving $24 - 12 = 12$ valid boy arrangements. Combining these, the total number of ways is $2 imes 6 imes 12 = 144$.