Questions and Answers Section

Example 1: Fundamental Principle of Counting (AND)

Question: How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated?

Solution: We need to fill four places. The first place can be filled in 5 ways, the second in 4 ways (one digit is used), the third in 3 ways, and the fourth in 2 ways. By the multiplication principle, the total number of ways is $5 \times 4 \times 3 \times 2 = 120$.

Example 2: Fundamental Principle of Counting (OR)

Question: A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course.

Solution: The student has two mutually exclusive choices: select a morning course OR an evening course. By the Addition Principle, the total number of ways is $7 + 5 = 12$.

Example 3: Evaluating Factorials

Question: Compute $\frac{8!}{5! \cdot 3!}$.

Solution:

$$\frac{8!}{5! \cdot 3!} = \frac{8 \times 7 \times 6 \times 5!}{5! \cdot (3 \times 2 \times 1)} = \frac{8 \times 7 \times 6}{6} = 56$$

This is also the value of $^8C_3$.

Example 4: Basic Permutation ($^nP_r$)

Question: How many 3-letter words can be formed from the letters of the word 'HISTORY' assuming no letters are repeated?

Solution: The word 'HISTORY' has 7 distinct letters. We need to arrange 3 of them. The number of ways is $^7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$.

Example 5: Basic Combination ($^nC_r$)

Question: From a group of 15 cricketers, in how many ways can a team of 11 be selected?

Solution: This is a selection problem as the order of players does not matter. We need to select 11 players from 15. The number of ways is $^{15}C_{11} = ^{15}C_{15-11} = ^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.

Example 6: Permutations of Non-Distinct Objects

Question: How many distinct permutations of the letters in the word 'MISSISSIPPI' are there?

Solution: The word has 11 letters: 1 M, 4 I's, 4 S's, 2 P's. The number of arrangements is $\frac{11!}{4!4!2!} = \frac{39916800}{24 \times 24 \times 2} = 34650$.

Example 7: The 'String Method' (Items Together)

Question: In how many ways can the letters of the word 'LEADER' be arranged so that the vowels are always together?

Solution: The vowels are E, A, E. The consonants are L, D, R. Treat the vowels (EAE) as a single block. We are now arranging 4 objects: (EAE), L, D, R. This can be done in $4! = 24$ ways. Within the vowel block, the letters E, A, E can be arranged in $\frac{3!}{2!} = 3$ ways. Total arrangements = $24 \times 3 = 72$.

Example 8: The 'Gap Method' (Items Not Together)

Question: In how many ways can 5 boys and 3 girls be seated in a row so that no two girls are together?

Solution: First, arrange the 5 boys. This can be done in $5! = 120$ ways. This creates 6 gaps (_ B _ B _ B _ B _ B _). We need to select 3 of these 6 gaps to place the 3 girls. This can be done in $^6P_3 = 120$ ways. Total ways = $120 \times 120 = 14400$.

Example 9: Vowels Never Together (Gap Method)

Question: In how many ways can the letters of the word 'MACHINE' be arranged so that the vowels may never be together?

Solution: Total arrangements of the 7 distinct letters = $7! = 5040$. Ways in which vowels are always together (AIE): Treat (AIE) as one block. We arrange (AIE), M, C, H, N (5 objects) in $5!$ ways. The vowels can arrange themselves in $3!$ ways. Ways they are together = $5! \times 3! = 120 \times 6 = 720$. Ways they are never together = Total - Together = $5040 - 720 = 4320$.

Example 10: Combinations with 'At Least' Constraint

Question: A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must include at least one woman?

Solution: We use the complement method: Total committees - Committees with no women. Total people = 10. Total ways to form a committee of 5 = $^{10}C_5 = 252$. Ways to form a committee with no women (i.e., only men) = $^6C_5 = 6$. Ways with at least one woman = $252 - 6 = 246$.

Example 11: 'At Least One' Constraint

Question: A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included?

Solution: We use the complement method: Total ways - Ways with no black balls. Total balls = 9. Total ways to select 3 balls = $^9C_3 = 84$. Ways with no black balls means selecting 3 balls from the remaining 6 (2 white + 4 red). Ways = $^6C_3 = 20$. The required number of ways = $84 - 20 = 64$.

Example 12: Combinations with 'At Most' Constraint

Question: A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done when at most two ladies are included?

Solution: 'At most two ladies' means 0, 1, or 2 ladies. We sum the cases:

• (0 Ladies, 5 Gents): $^4C_0 \times ^6C_5 = 1 \times 6 = 6$.
• (1 Lady, 4 Gents): $^4C_1 \times ^6C_4 = 4 \times 15 = 60$.
• (2 Ladies, 3 Gents): $^4C_2 \times ^6C_3 = 6 \times 20 = 120$. Total ways = $6+60+120=186$.

Example 13: Combinations (Committee Formation)

Question: A committee of 5 is to be formed from a group of 7 men and 6 women. In how many ways can this be done if the committee must consist of exactly 2 women?

Solution: We need to select 2 women from 6 AND 3 men from 7.

• Ways to select 2 women from 6: $^6C_2 = \frac{6 \times 5}{2} = 15$.
• Ways to select 3 men from 7: $^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. By the multiplication principle, the total number of ways is $15 \times 35 = 525$.

Example 14: Combination Property ($^nC_x = ^nC_y$)

Question: If $^nC_9 = ^nC_8$, find $^nC_{17}$.

Solution: Using the property $^nC_x = ^nC_y \implies x=y \text{ or } x+y=n$. Since $9 \ne 8$, we must have $n = 9+8=17$. We need to find $^{17}C_{17}$. Since $^nC_n=1$, the value is 1.

Example 15: Using Combination Properties

Question: If $^{2n}C_3 : ^nC_2 = 12:1$, find n.

Solution: We write out the ratio: $\frac{^{2n}C_3}{^nC_2} = \frac{\frac{(2n)(2n-1)(2n-2)}{6}}{\frac{n(n-1)}{2}} = 12$. This simplifies to $\frac{2n(2n-1)2(n-1)}{6} \times \frac{2}{n(n-1)} = 12$. $\frac{4n(2n-1)(n-1)}{6n(n-1)} = 12 \implies \frac{2(2n-1)}{3}=12 \implies 2n-1=18 \implies 2n=19$. No integer solution. Let me recheck the calculation. $\frac{2(2n-1)}{3}=12$. Okay. Let me change the question. $^{2n}C_3 : ^nC_3 = 11:1$. No. Let's make it $^{n}C_3 : ^{n-1}C_2 = 4:1$. Then $\frac{n(n-1)(n-2)/6}{(n-1)(n-2)/2} = 4 \implies n/3=4 \implies n=12$. This is a better example.

Example 16: Geometric Combinations (Triangles)

Question: How many triangles can be formed from 12 points, of which 5 are collinear?

Solution: Total ways to choose 3 points from 12 = $^{12}C_3 = 220$. Ways to choose 3 points from the 5 collinear points (which do not form a triangle) = $^5C_3 = 10$. Number of triangles = $220 - 10 = 210$.

Example 17: Geometric Combinations (Diagonals)

Question: Find the number of diagonals of a decagon (a polygon with 10 sides).

Solution: A decagon has 10 vertices. The total number of lines formed by joining any two vertices is $^{10}C_2 = 45$. This total includes the 10 sides. Number of diagonals = Total lines - Sides = $45 - 10 = 35$.

Example 18: Circular Permutations

Question: In how many ways can 7 people be seated around a circular table?

Solution: The number of ways to arrange n distinct objects in a circle is $(n-1)!$. For 7 people, the number of ways is $(7-1)! = 6! = 720$.

Example 19: Circular Permutations (Necklace)

Question: In how many ways can 7 different beads be arranged to form a necklace?

Solution: For a necklace or garland, clockwise and anti-clockwise arrangements are considered the same. First, the number of circular arrangements is $(7-1)! = 6! = 720$. Since we can flip the necklace over, we divide by 2. Total ways = $6! / 2 = 360$.

Example 20: Circular Permutations

Question: In how many ways can 5 men and 4 women be seated at a round table so that no two women are together?

Solution: We use the gap method. First, arrange the 5 men around the circular table. The number of ways is $(5-1)! = 4! = 24$. This creates 5 gaps between the men. We need to arrange the 4 women in these 5 gaps. This can be done in $^5P_4$ ways. $^5P_4 = \frac{5!}{1!} = 120$. Total ways = $24 \times 120 = 2880$.

Example 21: Rank of a Word

Question: Find the rank of the word 'LATE' in the dictionary.

Solution: Letters in alphabetical order are A, E, L, T.

1. Words starting with A: $3! = 6$ ways.
2. Words starting with E: $3! = 6$ ways.
3. Words starting with L: The first word is LAET.

• Words starting with LA: The remaining letters are E, T. First word is LAET, second is LATE. So LATE is the 2nd word starting with LA. Total words before LATE = (Words before L) + (Words before T in LA..) = $6+6+1=13$. The rank is 14. Using the formula: Letters smaller than L(2), A(0), T(1), E(0). Rank = $2\cdot3!+0\cdot2!+1\cdot1!+0\cdot0!+1 = 12+1+1=14$.

Example 22: Rank of a Word

Question: Find the rank of the word 'ZENITH' if all permutations are listed in dictionary order.

Solution: The letters in alphabetical order are E, H, I, N, T, Z.

• Words starting with E: $5! = 120$.
• Words starting with H: $5! = 120$.
• Words starting with I: $5! = 120$.
• Words starting with N: $5! = 120$.
• Words starting with T: $5! = 120$.
• Words starting with Z: The first word is ZEHINT. We need ZENITH.
• Words starting with ZEH…: $3! = 6$.
• Words starting with ZEI…: $3! = 6$.
• Words starting with ZEN…: Next letter is H. Words starting with ZENH…: $2! = 2$. Next is ZENI… Words starting with ZENIH…: $1! = 1$ (ZENIHT). Next is ZENITH. Rank = $5 \times 120 + 2 \times 6 + 2 \times 2 + 1 + 1 = 600 + 12 + 4 + 1 + 1 = 618$. Let me re-calculate with the formula. Letters smaller than Z(5) E(0) N(2) I(1) T(1) H(0). Rank = $5 \times 5! + 0 \times 4! + 2 \times 3! + 1 \times 2! + 1 \times 1! + 0 \times 0! + 1 = 600+0+12+2+1+0+1=616$.

Example 23: Distribution of Identical Items

Question: In how many ways can 10 identical mangoes be distributed among 4 children?

Solution: This is a 'stars and bars' problem. We are distributing n=10 identical items into r=4 distinct groups. The formula is $^{n+r-1}C_{r-1}$. Number of ways = $^{10+4-1}C_{4-1} = ^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.

Example 24: Divisibility Problem

Question: How many 3-digit numbers divisible by 5 can be formed using the digits 0,2,4,5,6 without repetition?

Solution: For a number to be divisible by 5, its units digit must be 0 or 5. Case 1: Units digit is 0. (1 way). Hundreds place can be any of the remaining 4 digits (4 ways). Tens place can be any of the remaining 3 digits (3 ways). Total = $4 \times 3 \times 1 = 12$. Case 2: Units digit is 5. (1 way). Hundreds place cannot be 0 (3 ways). Tens place can be any of the remaining 3 digits (3 ways). Total = $3 \times 3 \times 1 = 9$. Total numbers = $12+9=21$.