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Mixed Applications (Seating, Digits, Passwords, Multi-stage Problems)

Introduction

Mixed applications combine counting ideas - permutations, combinations, repetition rules, and conditional reasoning - into multi-step problems such as seating with constraints, forming passwords, or building numbers with digit rules.

This pattern is important because real exam questions rarely match a single formula: you must translate the problem into independent stages, pick the correct rule for each stage, then multiply or add case-results carefully.

Pattern: Mixed Applications (Seating, Digits, Passwords, Multi-stage Problems)

Pattern

Key idea: break the problem into clear stages (choose positions, apply constraints, count internal arrangements), use the appropriate counting rule for each stage (nPr, nCr, n^r), then combine stage-results by multiplication or addition.

Frequently used steps:

  • Stage decomposition: Identify independent choices (which seats, which last digit, which block placement).
  • Choose rule per stage: Use nPr when order matters, nCr when it doesn't, and nr when repetition is allowed.
  • Combine results: Multiply independent stage counts; add mutually exclusive case counts.
  • Sanity-check: Result must be ≤ corresponding unconstrained total (e.g., ≤ n! for permutations of n distinct items).

Step-by-Step Example

Question

How many 4-digit numbers can be formed from digits 0-9 if repetition is NOT allowed and the number must be divisible by 5?

Solution

  1. Step 1: Translate the divisibility condition into cases.

    A 4-digit number is divisible by 5 iff its last digit is 0 or 5. We'll count numbers with last digit 0 and with last digit 5, then add.

  2. Step 2: Case A - last digit = 0.

    First digit (thousands place) cannot be 0 (no leading zero) and cannot repeat digits. Available digits for first place = digits 1-9 → 9 choices. After fixing first and last, middle two places are filled from remaining 8 digits in order → permutations P(8,2) = 8 × 7 = 56.

    So count for Case A = 9 × 56 = 504.

  3. Step 3: Case B - last digit = 5.

    Last digit fixed as 5. First digit cannot be 0 or 5 → choices = digits {1..9} minus 5 → 8 choices. After choosing first and fixing last, remaining digits for middle two positions = 8, so middle positions count = P(8,2) = 8 × 7 = 56.

    So count for Case B = 8 × 56 = 448.

  4. Step 4: Combine mutually exclusive cases.

    Total valid numbers = Case A + Case B = 504 + 448 = 952.

  5. Final Answer:

    952 four-digit numbers satisfy the conditions.

  6. Quick Check:

    Unconstrained 4-digit numbers with no repetition = 9 × 9 × 8 × 7 = 4,536 (first digit 1-9, then any remaining). Our result 952 is less than 4,536 - plausible ✅

Quick Variations

1. Password-type: If repetition is allowed, use nr for independent positions (e.g., 104=10,000 for 4-digit with repetition).

2. Seating with blocks: For people who must sit together, treat them as one block then multiply by internal permutations.

3. Mixed digit constraints: If divisibility by 2 (even) + no repetition, split by last-digit parity cases similar to the example.

4. Multi-stage with selection + arrangement: Choose r items (nCr), then arrange them (r!) → equivalent to nPr.

Trick to Always Use

  • Step 1: Break the problem into independent stages or mutually exclusive cases (e.g., last digit choices, block vs single).
  • Step 2: For each stage, pick the right counting rule: nPr if order matters, nCr if order doesn't, nr if repetition allowed.
  • Step 3: Multiply counts from independent stages; add counts across exclusive cases.
  • Step 4: Run a quick sanity check: result should be ≤ the most relaxed total (e.g., ≤ n!, ≤ nr, or ≤ 2n depending on context).

Summary

Summary

  • Break every mixed problem into clear stages or mutually exclusive cases before counting.
  • Select the correct counting rule for each stage: nPr, nCr, or nr depending on order and repetition.
  • Multiply counts across independent stages and add results across mutually exclusive cases.
  • Always sanity-check the final count against the maximum unconstrained possibilities to ensure plausibility.

Example to remember:
Convert constraints into stages (like last-digit cases), apply rules per stage, then combine - exactly as done to reach 952 in the example.

Practice

(1/5)
1. How many 3-digit numbers can be formed using digits 1, 2, 3, 4, and 5 if repetition is allowed?
easy
A. 60
B. 125
C. 64
D. 120

Solution

  1. Step 1: Identify digits and rule.

    Available digits = 5 (1-5); repetition allowed → each of the 3 positions can be any of the 5 digits.

  2. Step 2: Apply the rule for repetition.

    Total = 53 = 125.

  3. Final Answer:

    125 → Option B.

  4. Quick Check:

    Each of the 3 places has 5 choices → 5 × 5 × 5 = 125 ✅
Hint: When repetition is allowed use n^r.
Common Mistakes: Using permutations (nPr) instead of n^r for repetition-allowed positions.
2. How many 4-letter words can be formed from the letters of the word 'SQUARE' if repetition is NOT allowed?
easy
A. 360
B. 720
C. 120
D. 240

Solution

  1. Step 1: Count available letters.

    'SQUARE' has 6 distinct letters.

  2. Step 2: Choose and arrange 4 letters (order matters, no repetition).

    Number = 6P4 = 6 × 5 × 4 × 3 = 360.

  3. Final Answer:

    360 → Option A.

  4. Quick Check:

    Multiply descending choices for each position: 6 × 5 × 4 × 3 = 360 ✅
Hint: Use nPr when order matters and repetition is not allowed.
Common Mistakes: Using combinations (nCr) instead of permutations (nPr) when order matters.
3. In how many ways can 5 people be seated in a row if A and B do not sit together?
easy
A. 96
B. 72
C. 120
D. 48

Solution

  1. Step 1: Total arrangements without restriction.

    Total = 5! = 120.

  2. Step 2: Count arrangements where A and B sit together.

    Treat A and B as one block → now 4 items → arrangements = 4! × 2! = 24 × 2 = 48.

  3. Step 3: Subtract to get 'not together'.

    Not together = Total - Together = 120 - 48 = 72.

  4. Final Answer:

    72 → Option B.

  5. Quick Check:

    Together (48) + Not together (72) = 120 (total) ✅
Hint: Compute Total - Together; treat the pair as a block to count 'together'.
Common Mistakes: Forgetting the internal 2! for the paired block.
4. How many 3-digit even numbers can be formed using digits 1 to 7 without repetition?
medium
A. 90
B. 120
C. 150
D. 100

Solution

  1. Step 1: Fix the units (last) digit to be even.

    Even digits from 1-7 are 2, 4, 6 → 3 choices for the units place.

  2. Step 2: Choose the hundreds (first) digit.

    First digit cannot be zero (not present) and cannot be the chosen last digit → from remaining 6 digits → 6 choices.

  3. Step 3: Choose the tens digit.

    Remaining digits now = 5 choices.

  4. Step 4: Multiply choices across positions.

    Total = 3 × 6 × 5 = 90.

  5. Final Answer:

    90 → Option A.

  6. Quick Check:

    Count by fixing last digit first (3 ways) then fill hundreds and tens without repetition → 3 × 6 × 5 = 90 ✅
Hint: Fix constrained positions (last digit even) first, then count remaining positions without repetition.
Common Mistakes: Counting leading zeros or not enforcing non-repetition when filling remaining places.
5. A 4-digit password is made of digits 0-9 where repetition is allowed but the password must start with a non-zero digit. How many passwords can be formed?
medium
A. 10,000
B. 9,000
C. 9,999
D. 9,090

Solution

  1. Step 1: Handle first-digit restriction.

    First digit must be 1-9 → 9 choices.

  2. Step 2: Remaining digits.

    Each of the remaining 3 positions can be any of 10 digits (0-9) with repetition allowed → 10 × 10 × 10.

  3. Step 3: Multiply all stages.

    Total = 9 × 10 × 10 × 10 = 9,000.

  4. Final Answer:

    9,000 → Option B.

  5. Quick Check:

    All 4-digit numbers without leading zero = 9,000 (from 1,000 to 9,999) ✅
Hint: Treat the restricted first digit separately, then multiply by choices for remaining positions.
Common Mistakes: Allowing a leading zero (which would produce 3-digit numbers) or forgetting repetition rules.

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