Mixed Permutation and Combination

[MathDefinition title="Mixed Problems"] Mixed problems involve using both permutations and combinations in the same problem. The key is to identify when order matters (permutation) and when it doesn't (combination). [/MathDefinition]

[MathDefinition title="Problem Solving Strategy"]

1. Identify what needs to be selected (combination)
2. Identify what needs to be arranged (permutation)
3. Apply multiplication principle
4. Consider special cases and restrictions [/MathDefinition]

[MathKeyFormula title="Key Concepts"]

1. Selection: ⁿCᵣ (order doesn't matter)
2. Arrangement: ⁿPᵣ or r! (order matters)
3. Multiplication principle: If task A can be done in m ways and task B in n ways, then both can be done in m × n ways
4. Addition principle: If task A can be done in m ways and task B in n ways, and they are mutually exclusive, then either can be done in m + n ways [/MathKeyFormula]

[MathNote title="Identifying Permutation vs Combination"] Ask yourself: "Does the order of selection matter?"

• YES → Use permutation
• NO → Use combination
• First select (combination), then arrange (permutation) [/MathNote]

[MathExample title="Example 1: Selection and Arrangement"] Problem: How many 4-digit numbers can be formed from digits 1, 2, 3, 4, 5, 6, 7 such that digits do not repeat and the number is even?

Solution: Step 1: Select the last digit (must be even: 2, 4, or 6) Ways to choose last digit = ³C₁ = 3

Step 2: Select the remaining 3 digits from the remaining 6 digits Ways to choose 3 digits = ⁶C₃ = 20

Step 3: Arrange the 3 selected digits in the first three positions Ways to arrange = 3! = 6

Total numbers = 3 × 20 × 6 = 360

Therefore, 360 such numbers can be formed. [/MathExample]

[MathExample title="Example 2: Committee with Roles"] Problem: A committee of 3 people is to be selected from 5 men and 4 women. If the committee must have at least 1 woman and 1 man, and one person is designated as chairperson, how many ways can the committee be formed?

Solution: Case 1: 1 woman, 2 men Select 1 woman: ⁴C₁ = 4 Select 2 men: ⁵C₂ = 10 Select chairperson from 3 members: ³C₁ = 3 Total for Case 1 = 4 × 10 × 3 = 120

Case 2: 2 women, 1 man Select 2 women: ⁴C₂ = 6 Select 1 man: ⁵C₁ = 5 Select chairperson from 3 members: ³C₁ = 3 Total for Case 2 = 6 × 5 × 3 = 90

Total committees = 120 + 90 = 210

Therefore, 210 such committees can be formed. [/MathExample]

[MathExample title="Example 3: Arranging with Restrictions"] Problem: In how many ways can the letters of the word 'ARTICLE' be arranged so that vowels occupy even positions?

Solution: The word ARTICLE has 7 letters: A, R, T, I, C, L, E Vowels: A, I, E (3 vowels) Consonants: R, T, C, L (4 consonants)

Even positions: 2, 4, 6 (3 positions) Odd positions: 1, 3, 5, 7 (4 positions)

Step 1: Arrange 3 vowels in 3 even positions Ways = ³P₃ = 3! = 6

Step 2: Arrange 4 consonants in 4 odd positions Ways = ⁴P₄ = 4! = 24

Total arrangements = 6 × 24 = 144

Therefore, 144 such arrangements are possible. [/MathExample]

[MathExample title="Example 4: Selection with Conditions"] Problem: From 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Solution: Step 1: Select 3 consonants from 7 Ways = ⁷C₃ = 35

Step 2: Select 2 vowels from 4 Ways = ⁴C₂ = 6

Step 3: Arrange the 5 selected letters Ways = 5! = 120

Total words = 35 × 6 × 120 = 25,200

Therefore, 25,200 such words can be formed. [/MathExample]