Comprehensive Study Guide: Permutations and Combinations

Geometric Applications of Combinations and Diagonals

In the study of polygons and sets of points on a plane, combinations are used to determine the count of lines, triangles, and diagonals. For a polygon or a plane containing $n$ points where no three points are collinear:

The number of straight lines that can be formed by connecting any two of the $n$ points is given by the formula: $\text{Lines} = {^nC_2}$

The number of diagonals in a polygon with $n$ vertices is the total number of lines minus the number of sides: $\text{Diagonals} = {^nC_2} - n$

Example 1: For a 16-sided polygon ($n = 16$): $\text{Total lines} = {^{16}C_2} = \frac{16 \times 15}{2 \times 1} = 120$ $\text{Number of Diagonals} = 120 - 16 = 104$

Example 2: For 7 points on a plane ($n = 7$) where no two points are collinear: (i) Number of straight lines: ${^7C_2} = \frac{7 \times 6}{2 \times 1} = 21$ (ii) Number of triangles: ${^7C_3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ (iii) The number of diagonals for a 7-sided plane figure formed by these points: ${^7C_2} - 7 = 21 - 7 = 14$

Distribution and Grouping Principles

When dividing a group of items or people into equal-sized subgroups, specific permutation and combination rules apply.

To divide 6 students into two equal groups of 3 students each, the calculation follows the formula for partitioning: $\text{Ways} = \frac{6!}{3! \times 3! \times 2!}$ Note that the $2!$ is included because the order of the two resulting groups does not matter (they are indistinguishable sets).

Calculations for arrangements involving fixed positions: In a team of 14 members, if we need to form an 11-member team while keeping one specific captain fixed, the choice is made from the remaining 13 members to fill the remaining 10 spots: $\text{Selection} = {^{13}C_{10}}$

If we want to arrange 10 members in a line from the 14 available members, the formula is: $\text{Arrangements} = {^{14}C_{10}} \times 10!$

To invite one or more members out of 14 members to a party, the total number of ways is: $\text{Invitations} = 2^{14} - 1$

Fundamental Identities and Algebraic Properties

The following mathematical identities are established for the objects $n$ and $r$:

1. The relationship between combinations and factorials: ${^nC_r} = \frac{n!}{r!\,(n - r)!}$

2. The relationship involving the complement: ${^nC_r} = {^nC_{n - r}}$

3. The relationship between Permutations and Combinations: ${^nP_r} = r! \times {^nC_r}$

4. Algebraic evaluation of $n$: If ${^nC_{12}} = {^nC_8}$, then according to the property ${^nC_x} = {^nC_y} \implies x + y = n$, we find: $n = 12 + 8 = 20$

If ${^nC_x} = {^nC_y}$ where $x \neq y$, then $n = x + y$.

Word Permutations with Identical Letters

When arranging letters of a word, if certain letters are repeated, the total number of permutations is divided by the factorials of the counts of the repeated letters.

FATHER: Total letters = 6 (all distinct). Total permutations = $6! = 720$. Total re-arrangements (excluding the word itself) = $720 - 1 = 719$.

DHAKA: Total letters = 5, where 'A' is repeated twice. Total arrangements = $\frac{5!}{2!} = \frac{120}{2} = 60$. Total re-arrangements = $60 - 1 = 59$. Selection of 4 letters from DHAKA (letters are D, H, K, A, A):

• 4 distinct letters: ${^4C_4} = 1$

• 2 A's and 2 from 3 distinct: ${^2C_2} \times {^3C_2} = 1 \times 3 = 3$ Total combinations = $1 + 3 = 4$.

ADMISSION: Total letters = 9 (S repeated twice, I repeated twice). Total arrangements = $\frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720$.

BANGLADESH with 'DESH' fixed: Since 'DESH' stays in its position, we only arrange the remaining letters: B, A, N, G, L, A. Total letters to arrange = 6, with 'A' repeated twice. Permutations = $\frac{6!}{2!} = 360$.

RAJSHAHI and BARISAL Comparison: RAJSHAHI: 8 letters (A repeats twice, H repeats twice, I repeats twice - note correct count from transcript logic implies A and H repeats are primary). Transcript calculation for RAJSHAHI: $\frac{8!}{2! \times 2!} = 10080$. BARISAL: 7 letters (A repeats twice). Permutations = $\frac{7!}{2!} = 2520$. Ratio: $k = \frac{10080}{2520} = 4$.

MATHEMATICS: Total letters = 11 (M repeats twice, A repeats twice, T repeats twice). Condition: 'T' at first and last positions. Remaining letters to arrange: M, A, H, E, M, A, I, C, S (9 letters). Distinct repeats in remaining: M (2), A (2). Permutations = $\frac{9!}{2! \times 2!} = 90720$.

CALCULUS: Total letters = 8 (C repeats twice, U repeats twice, L repeats twice). Condition: 'U' at first and last positions. Remaining letters: C, A, L, C, L, S (6 letters). Repeats: C (2), L (2). Permutations = $\frac{6!}{2! \times 2!} = 180$.

Constraints: Vowels Together and Not Together

ENGINEERING: Total letters = 11 (E appears 3 times, N 3 times, G 2 times, I 2 times). Condition: All 'E's must be together. Treat 3 'E's as a single block. Total items for arrangement: (11 - 3 + 1) = 9. Items: (EEE), N, N, N, G, G, I, I, R. Arrangements = $\frac{9!}{3! \times 2! \times 2!} = 15120$.

SCIENCE: Total letters = 7 (C appears twice, E appears twice). Vowels (I, E, E) = 3 total, with E twice. Condition: Vowels together. Treat vowels (IEE) as one block. Remaining: S, C, N, C (4 letters). Total items = 4 + 1 = 5. Arrangements of items = $\frac{5!}{2!} = 60$ (for repeated C). Vowel block internal arrangement = $\frac{3!}{2!} = 3$. Total = $60 \times 3 = 180$.

TRIANGLE: Total letters = 8 (all distinct). Vowels = 3 (I, A, E). Total arrangements = $8! = 40320$. Vowels together: (8 - 3 + 1)! x 3! = $6! \times 3! = 720 \times 6 = 4320$. Vowels NOT together = $40320 - 4320 = 36000$.

Numerical Permutations and Constraints

Problem 38: Forming 4-digit numbers from {2, 3, 4, 5, 6, 7, 8} without repetition. Total digits $n = 7$, required $r = 4$. $^7P_4 = 7 \times 6 \times 5 \times 4 = 840$.

Problem 39: Three-digit odd numbers from {4, 5, 6, 7, 8}. Odd digits available for the units place: {5, 7}. Case 1: Unit place is 5. Remaining 2 places filled by 4 digits: $^4P_2 = 12$. Case 2: Unit place is 7. Remaining 2 places filled by 4 digits: $^4P_2 = 12$. Total odd numbers = $12 + 12 = 24$.

Problem 40: Three-digit numbers > 200 from {1, 3, 5, 7, 9}. First digit must be from {3, 5, 7, 9} (4 choices). Remaining 2 places from 4 remaining digits: $^4P_2 = 12$. Total numbers = $4 \times 12 = 48$.

Problem 41: Numbers > 4000 from {0, 3, 5, 6, 8} without repetition. Case 1: 4-digit numbers. First digit must be {5, 6, 8} (3 choices). Remaining 3 spots from 4 digits: $3 \times {^4P_3} = 3 \times 24 = 72$. Case 2: 5-digit numbers. First digit cannot be 0 (4 choices). Remaining 4 spots: $4 \times 4! = 4 \times 24 = 96$. Total numbers = $72 + 96 = 168$.

Advanced Word Problems

COURAGE: Total letters = 7 (Vowels: O, U, A, E - Consonants: C, R, G).

1. Condition: Starts with a vowel. First place: $^4P_1 = 4$. Remaining 6 places: $6! = 720$. Total = $4 \times 720 = 2880$.

2. Condition: Starts with a consonant. First place: $^3P_1 = 3$. Remaining 6 places: $6! = 720$. Total = $3 \times 720 = 2160$.

PERMUTATION: Condition: Vowels (e, u, a, i, o) do not change their relative positions. There are 5 vowels and 6 consonants (p, r, m, t, t, n). Consonants have 't' twice. We only arrange the consonants in their available spots: $\frac{6!}{2!} = 360$. Re-arrangement = $360 - 1 = 359$.

INSTITUTE: Total letters = 9 (I: 2, T: 3). Vowels = 3, Consonants = 6. Condition: Vowels occupy even positions. Places marked 1, 2, 3, 4, 5, 6, 7, 8, 9. Even places are 2, 4, 6, 8 (4 spots). Vowels (I, I, E) in 4 spots: $\frac{^4P_3}{2!} = 12$. Consonants (N, S, T, T, T) in remaining 6 spots: $\frac{6!}{3!} = 120$. Total = $12 \times 120 = 1440$.

EQUATION: Total letters = 8 (Vowels = 5: e, u, a, i, o; Consonants = 3: q, t, n). Condition: Consonants occupy odd positions. Odd places: 1, 3, 5, 7 (4 spots). Arranging 3 consonants in 4 spots: $^4P_3 = 24$. Arranging 5 vowels in remaining 5 spots: $5! = 120$. Total = $24 \times 120 = 2880$.