Permutations And Combinations – JEE Main MathematicsPractice Questions & PYQs

The number of triangles formable from the vertices of an $n$-sided polygon is given by the combination $\binom{n}{3}$. By substituting $T_{n+1} = \binom{n+1}{3}$ and $T_n = \binom{n}{3}$ into the equation $T_{n+1} - T_n = 10$, we apply Pascal's Identity to simplify the expression to $\binom{n}{2} = 10$. Solving the resulting equation $\frac{n(n-1)}{2} = 10$ provides the integer value for $n$.

For groups of identical items, the number of ways to choose zero or more items is $n+1$, where $n$ is the count of items in the group. We multiply the possibilities for the white, green, and black balls, which is $(10+1) \times (9+1) \times (7+1)$, and since the condition requires selecting at least one ball, we subtract 1 from this product to exclude the single case where no balls are chosen from any category.

Distributing 10 identical balls into 4 distinct boxes such that each box receives at least one ball is equivalent to placing 3 dividers in the 9 spaces between the 10 balls, calculated using ${}^{9}C_{3}$. Separately, the number of ways to select 3 items from a set of 9 is a standard combination problem calculated as ${}^{9}C_{3}$. While both statements describe mathematically equivalent expressions, the logic behind the stars and bars method for distribution is conceptually independent of the simple selection of positions, meaning the second statement does not serve as an explanation for the first.

To form a triangle, we need to select 3 non-collinear points from the available set. Total ways to pick 3 points from 10 is given by ${}^{10}C_3$, but since 6 points are collinear, they cannot form a triangle, so we must subtract the number of ways to pick 3 points from those 6, which is ${}^{6}C_3$. Calculating these yields $120 - 20$, resulting in exactly 100 possible triangles.

Since we need to perform two independent selections of balls from each urn simultaneously, the total number of ways is the product of the individual selection possibilities. Selecting 2 balls from 3 in urn A gives ${}^{3}C_2$ ways, and selecting 2 balls from 9 in urn B gives ${}^{9}C_2$ ways. Multiplying these values, $3 imes 36$, provides 108 distinct ways to perform the transfer.

Selecting 4 novels from 6 is ${}^{6}C_4$ and 1 dictionary from 3 is ${}^{3}C_1$, giving a total of $15 imes 3 = 45$ selection combinations. Once selected, the dictionary must occupy the fixed middle position, leaving the 4 novels to be arranged in the remaining 4 spots in $4!$ ways. Multiplying the selection combinations by the arrangement factor, $45 imes 24$, results in 1080 possible arrangements.

Choosing 6 ice creams from 5 distinct types is a combination with repetition problem, solved using the formula ${}^{n+r-1}C_{r-1}$ where $n=6$ and $r=5$, yielding ${}^{10}C_4$. Since the value of ${}^{10}C_4$ is 210, Statement 1 claiming the answer is ${}^{10}C_5$ is false. Statement 2, which refers to the arrangement of 6 A's and 4 B's, is calculated as rac{10!}{6!4!}, which is equivalent to ${}^{10}C_4$, making the second statement true.

The gap method provides the framework for arranging items with non-adjacency constraints, requiring the placement of non-constrained letters first. The 7 non-'S' letters (M, I, I, I, I, P, P) can be arranged in $\frac{7!}{4!2!} = 105$ ways, which corresponds to $7 \cdot {}^6C_4$. These letters create 8 distinct gaps, and choosing 4 of these gaps to place the four 'S' letters is done in ${}^8C_4$ ways, leading to the total expression $7 \cdot {}^6C_4 \cdot {}^8C_4$.

Partitioning a set of 12 distinct elements into three labeled subsets of equal size 4 is calculated using the multinomial coefficient $\frac{12!}{4!4!4!}$. This logic follows from selecting 4 elements for the first set from 12, then 4 for the second from the remaining 8, and the final 4 for the third from the remaining 4, which equates to $\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = \frac{12!}{4!8!} \times \frac{8!}{4!4!} \times 1 = \frac{12!}{(4!)^3}$.

The total number of ways a voter can choose candidates is found by calculating the sum of combinations for each allowed number of votes from 1 to 4. Applying the addition rule for these mutually exclusive cases, we calculate ${}^{10}C_1 + {}^{10}C_2 + {}^{10}C_3 + {}^{10}C_4$. Substituting the combinatorial values $10 + 45 + 120 + 210$ results in a total of 385 ways.

Lexicographical order in a dictionary is determined by the alphabetical position of letters, with ranks calculated by summing the permutations of preceding letter combinations. For the word SACHIN, letters are ordered A, C, H, I, N, S. Counting arrangements that start with letters before 'S', we find $5 \cdot 5! = 600$ words, and since the target word follows these combinations immediately, we add 1 to reach a rank of 601.

The number of arrangements with specific relative constraints on subsets of items is found by dividing the total number of permutations by the number of ways those items can be arranged. Since the word GARDEN has 6 distinct letters, there are $6! = 720$ total permutations, and because the two vowels, A and E, can appear in either the relative order AE or EA, dividing by $2!$ ensures A always precedes E, yielding 360 valid arrangements.

Distributing n identical objects into k distinct bins such that no bin is empty is solved using the stars and bars method, specifically the formula $\binom{n-1}{k-1}$. In this case, there are 8 balls and 3 boxes, meaning the calculation becomes $\binom{8-1}{3-1}$, which simplifies to $\binom{7}{2}$. Evaluating this combination as $\frac{7 \times 6}{2 \times 1}$ results in 21 possible ways.

The expression can be evaluated by decomposing the middle term $2 \binom{n}{r}$ into $\binom{n}{r} + \binom{n}{r}$ to apply Pascal's identity, which states $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$. Rearranging the original terms allows for grouping as $(\binom{n}{r+1} + \binom{n}{r}) + (\binom{n}{r} + \binom{n}{r-1})$, which simplifies to $\binom{n+1}{r+1} + \binom{n+1}{r}$. Applying the identity once more results in the final form $\binom{n+2}{r+1}$.

Selecting 10 questions out of 13 with the requirement of at least 4 from the first 5 leads to two mutually exclusive scenarios. The first involves choosing 4 from the first group and 6 from the remaining 8, calculated as $\binom{5}{4} \times \binom{8}{6} = 5 \times 28 = 140$. The second scenario involves choosing all 5 from the first group and 5 from the remaining 8, calculated as $\binom{5}{5} \times \binom{8}{5} = 1 \times 56 = 56$. Summing these results gives a total of $140 + 56 = 196$ choices.

To seat 6 men and 5 women at a round table with no two women adjacent, first seat the 6 men in $(6-1)! = 5!$ ways. This arrangement leaves 6 distinct gaps between the men, into which the 5 women can be seated in $P(6,5)$ ways, which is equivalent to $6!$. Multiplying the arrangement of the men by the possible positions for the women results in $5! \times 6!$ total ways.

Constructing a four-digit odd number requires that the thousands place is not zero, the middle digits can be any of the 6 available values, and the units place is restricted to the odd digits {1, 3, 5, 7}. Applying the counting principle yields $5 \times 6 \times 6 \times 4 = 720$ distinct possibilities.

Divisibility by 3 is determined by the sum of a number's digits being a multiple of 3. Using the set {0, 1, 2, 3, 4, 5}, we identify two valid 5-digit subsets: {1, 2, 3, 4, 5} sums to 15, allowing $5! = 120$ arrangements, and {0, 1, 2, 4, 5} sums to 12, allowing $4 imes 4! = 96$ arrangements because zero cannot occupy the first position. Combining these gives a total of 216 numbers.

A four-digit number between 1000 and 4000 requires the thousands digit to be 1, 2, or 3, while the remaining positions can be any of the 5 digits since repetition is allowed. Calculating the total possibilities gives $3 imes 5 imes 5 imes 5 = 375$, and subtracting the case where the number equals 1000 results in 374 valid options.

The sum of integers between 1 and 100 divisible by 2 or 5 is determined by the inclusion-exclusion principle, adding the sum of multiples of 2 and the sum of multiples of 5, then subtracting the sum of multiples of 10 to avoid overcounting. The sum of multiples of 2 is an arithmetic progression with 50 terms, calculated as $\frac{50}{2}(2 + 100) = 2550$. The sum of multiples of 5 has 20 terms, $\frac{20}{2}(5 + 100) = 1050$, while the sum of multiples of 10 has 10 terms, $\frac{10}{2}(10 + 100) = 550$. Performing the calculation $2550 + 1050 - 550$ yields 3050.