Permutation and Combination Concepts

1. Restricted Letter Arrangements

• Formula: For (n) distinct objects:

$$n!$$

If (k) vowels must be together, treat all vowels as one block:

$$(n-k+1)!\times k!$$

Adjust for repeated vowels by dividing by their factorials.

• Example: In how many ways can the letters of MANAGER be arranged so that all vowels are together? Solution: Vowels are (A,A,E).

Treat (AAE) as one block:

$$(AAE),M,N,G,R$$

So there are 5 units:

$$5!$$

The vowels can be arranged internally in:

$$\frac{3!}{2!}=3$$

Therefore:

$$5!\times\frac{3!}{2!} =120\times3 =\boxed{360}$$

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2. Vowels Never Together

• Formula:

$$\text{Required arrangements} =\text{Total arrangements}-\text{Vowels together arrangements}$$

For distinct vowels, another useful method is to arrange consonants first and place vowels in the gaps.

• Example: In how many ways can the letters of APPLE be arranged so that the two vowels are not together? Solution: Letters: (A,P,P,L,E).

Total arrangements:

$$\frac{5!}{2!}=60$$

Vowels together: Treat (AE) or (EA) as one block.

Units:

$$(AE),P,P,L$$

Arrangements:

$$\frac{4!}{2!}\times2!=24$$

Therefore:

$$60-24=\boxed{36}$$

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3. Mixed Selection & Arrangement

• Formula: Selection of (r) objects from (n):

$${}^nC_r=\frac{n!}{r!(n-r)!}$$

Arrangement of (r) objects:

$${}^nP_r=\frac{n!}{(n-r)!}$$

If selection is followed by arrangement:

$${}^nC_r\times r!={}^nP_r$$

• Example: In how many ways can a President, Vice-President and Secretary be selected from 8 people? Solution: The three positions are distinct, so order matters:

$${}^8P_3 =\frac{8!}{5!}$$ $$=8\times7\times6 =\boxed{336}$$

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4. Selection of a Committee

• Formula: When order does not matter:

$${}^nC_r=\frac{n!}{r!(n-r)!}$$

• Example: How many committees of 3 can be formed from 8 people? Solution:

$${}^8C_3 =\frac{8!}{3!5!}$$ $$=\frac{8\times7\times6}{3\times2\times1} =\boxed{56}$$

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5. Identical Object Arrangements

• Formula: If (n) objects contain (p,q,r,\ldots) identical objects:

$$\frac{n!}{p!q!r!\cdots}$$

• Example: How many distinct arrangements can be made using the letters of SUCCESS? Solution: There are 7 letters:

$$S=3,\quad C=2,\quad U=1,\quad E=1$$

Therefore:

$$\frac{7!}{3!2!} =\frac{5040}{12} =\boxed{420}$$

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6. Circular Arrangements

• Formula: Number of arrangements of (n) distinct people around a circle:

$$(n-1)!$$

If clockwise and anticlockwise arrangements are considered identical:

$$\frac{(n-1)!}{2}$$

• Example: In how many ways can 6 people sit around a circular table? Solution: Fix one person and arrange the remaining 5:

$$(6-1)!=5!$$ $$=\boxed{120}$$

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7. Arrangement with Two Particular Objects Together

• Formula: Treat the two objects as one block:

$$(n-1)!\times2!$$

for distinct objects.

• Example: In how many ways can 5 people A, B, C, D and E stand in a row if A and B must stand together? Solution: Treat (AB) as one block.

Units:

$$(AB),C,D,E$$

Arrange 4 units:

$$4!$$

A and B can switch places:

$$2!$$

Therefore:

$$4!\times2!=24\times2=\boxed{48}$$

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8. Arrangement with Two Particular Objects Not Together

• Formula:

$$\text{Required}=\text{Total arrangements}-\text{Together arrangements}$$

• Example: In how many ways can 5 people A, B, C, D and E stand in a row if A and B cannot stand together? Solution: Total:

$$5!=120$$

A and B together:

$$4!\times2!=48$$

Therefore:

$$120-48=\boxed{72}$$

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9. Fixed Position Arrangements

• Formula: Fix the required object/objects first, then arrange the remaining objects:

$$(n-k)!$$

when (k) distinct positions are fixed.

• Example: How many arrangements of the letters A, B, C, D, E have A in the first position? Solution: Fix A in the first position.

Arrange B, C, D, E:

$$4!=\boxed{24}$$

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10. Restricted Position / Not in Original Position

• Formula: For (n) objects where no object can occupy its original position, use derangements:

$$!n=n!\left(1-\frac1{1!}+\frac1{2!}-\frac1{3!}+\cdots+\frac{(-1)^n}{n!}\right)$$

For small (n):

$$!3=2,\quad !4=9,\quad !5=44$$

• Example: In how many ways can 4 letters be placed into 4 addressed envelopes so that no letter goes into the correct envelope? Solution: This is a derangement of 4 objects:

$$!4=\boxed9$$

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11. Dictionary Rank Calculations

• Formula:

$$\text{Rank}=\text{Number of words before the given word}+1$$

At each position, count unused letters alphabetically smaller than the current letter and multiply by the factorial of the remaining positions.

• Example: Find the rank of MATH in dictionary order. Solution: Letters in alphabetical order:

$$A,H,M,T$$

Before words beginning with M:

• Starting with A: (3!=6)
• Starting with H: (3!=6)

After fixing M, the next letter is A. No unused letter is smaller than A.

Before MATH with prefix MA:

• Remaining letters H,T.
• H comes before T, giving:

$$2!=2$$

Total words before:

$$6+6+0=12$$

Therefore:

$$\text{Rank}=12+1=\boxed{13}$$

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12. Rank of Words with Repeated Letters

• Formula: For repeated letters, count distinct permutations at each stage:

$$\frac{n!}{a!b!\cdots}$$

where (a,b,\ldots) are multiplicities of repeated letters.

• Example: Find the rank of BANANA. Solution: Letters:

$$A^3,N^2,B$$

Words before BANANA:

First letter B: no smaller letter exists.

Second letter A: no smaller unused letter.

Third letter N: available smaller letter is A.

Fix BA and place A: Remaining (A,A,N):

$$\frac{3!}{2!}=3$$

Thus 3 words occur before BANANA.

Therefore:

$$\boxed{4}$$

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13. Alternating Men and Women

• Formula: If (n) men and (n) women must alternate:

$$n!\times n!$$

If there are (n) men and (n+1) women, the women must occupy both ends:

$$(n+1)!\times n!$$

• Example: In how many ways can 4 men and 4 women sit in a row alternately? Solution: Two possible patterns:

$$M-W-M-W-M-W-M-W$$

Arrange men:

$$4!$$

Arrange women:

$$4!$$

Total:

$$4!\times4!=576$$

There are two starting patterns, so:

$$2\times576=\boxed{1152}$$

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14. Repetition Allowed in Arrangements

• Formula: If (n) choices are available for each of (r) positions and repetition is allowed:

$$n^r$$

• Example: How many 4-digit PINs can be formed using digits 0–9 if repetition is allowed? Solution: Each position has 10 choices:

$$10\times10\times10\times10$$ $$=\boxed{10^4=10000}$$

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15. Arrangements Without Repetition

• Formula:

$${}^nP_r=\frac{n!}{(n-r)!}$$

• Example: How many 3-digit numbers can be formed from (1,2,3,4,5) without repetition? Solution:

$${}^5P_3 =5\times4\times3 =\boxed{60}$$

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16. Number Formation with Zero

• Formula: For an (r)-digit number, the first digit cannot be zero. If there are (n) available digits including 0 and no repetition:

$$(n-1)\times{}^{n-1}P_{r-1}$$

• Example: How many 3-digit numbers can be formed using (0,1,2,3,4) without repetition? Solution: First digit: 4 choices ((1,2,3,4)).

Second digit: 4 choices.

Third digit: 3 choices.

Therefore:

$$4\times4\times3=\boxed{48}$$

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Advanced Variants

17. Vowels Together and Consonants Together

• Formula: Treat each required group as a block, then arrange the blocks and multiply by internal arrangements.
• Example: In how many ways can the letters of AEBCD be arranged if all vowels are together and all consonants are together? Solution: Vowels: (A,E) → one block.

Consonants: (B,C,D) → one block.

Arrange the two blocks:

$$2!$$

Arrange vowels:

$$2!$$

Arrange consonants:

$$3!$$

Total:

$$2!\times2!\times3! =\boxed{24}$$

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18. At Least One Particular Object Selected

• Formula:

$$\text{At least one}=\text{Total}-\text{None}$$

• Example: From 8 people, how many committees of 3 can be formed containing at least one of A and B? Solution: Total committees:

$${}^8C_3=56$$

Committees containing neither A nor B:

$${}^6C_3=20$$

Therefore:

$$56-20=\boxed{36}$$

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19. Particular Objects Must Be Separated

• Formula: Arrange the unrestricted objects first. Use the gaps between them to place the restricted objects.
• Example: In how many ways can 3 identical vowels be placed among 4 distinct consonants so that no two vowels are together? Solution: Arrange the 4 consonants:

$$4!$$

They create 5 gaps:

_,C,_,C,_,C,_,C,_

Choose 3 of the 5 gaps:

$${}^5C_3=10$$

Total:

$$4!\times10=\boxed{240}$$

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20. Committee with Required and Excluded Members

• Formula: If (k) specific people must be included and (m) specific people must be excluded from a committee of (r) chosen from (n):

$${}^{n-k-m}C_{r-k}$$

• Example: From 10 people, how many committees of 4 can be formed if A must be included and B must be excluded? Solution: A is already selected.

B is unavailable.

Choose remaining 3 from the other 8:

$${}^8C_3 =\boxed{56}$$

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21. Circular Arrangement with Two People Together

• Formula: Treat the required pair as one block:

$$(n-2)!\times2!$$

for (n) distinct people around a circle.

• Example: In how many ways can 6 people sit around a circular table if A and B must sit together? Solution: Treat A and B as one block.

Total units:

$$5$$

Circular arrangements:

$$(5-1)!=4!$$

A and B can switch:

$$2!$$

Therefore:

$$4!\times2=\boxed{48}$$

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22. Identical Objects Distributed into Distinct Groups

• Formula: Number of ways to distribute (n) identical objects among (r) distinct groups when empty groups are allowed:

$${}^{n+r-1}C_{r-1}$$

When every group must receive at least one:

$${}^{n-1}C_{r-1}$$

• Example: In how many ways can 7 identical balls be distributed among 3 distinct boxes if a box may be empty? Solution: Using stars and bars:

$${}^{7+3-1}C_{3-1} ={}^{9}C_2$$ $$=\frac{9\times8}{2} =\boxed{36}$$