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

Introduction

Mixed applications में counting ideas - permutations, combinations, repetition rules और conditional reasoning - को एक साथ कई-step वाले problems में उपयोग किया जाता है, जैसे seating with constraints, passwords बनाना या digit-rules के साथ numbers तैयार करना।

यह pattern इसलिए महत्वपूर्ण है क्योंकि असली exam questions शायद ही किसी एक formula पर आधारित हों: आपको problem को independent stages में तोड़ना होता है, हर stage के लिए सही rule चुनना होता है, और फिर case-results को सही तरह जोड़ना या गुणा करना होता है।

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

Pattern

मुख्य विचार: problem को clear stages में तोड़ें (positions चुनना, constraints लगाना, internal arrangements count करना), हर stage के लिए सही counting rule (nPr, nCr, nr) लागू करें, और फिर stage-results को multiply या add करें।

Frequently used steps:

  • Stage decomposition: Independent choices पहचानें (कौन-सी seats, कौन-सा last digit, कौन-सा block placement)।
  • Choose rule per stage: Order matter करे तो nPr, order न matter करे तो nCr, repetition allowed हो तो nr।
  • Combine results: Independent stages → multiply; mutually exclusive cases → add।
  • Sanity-check: Result संबंधित unconstrained total (जैसे n!) से ≤ होना चाहिए।

Step-by-Step Example

Question

Digits 0-9 से repetition allowed नहीं है और number 5 से divisible होना चाहिए। ऐसे कितने 4-digit numbers बनाए जा सकते हैं?

Solution

  1. Step 1: divisibility condition को cases में बदलें।

    एक 4-digit number तभी 5 से divisible होगा जब उसका last digit 0 या 5 हो। इसलिए दो cases गिनेंगे: last digit 0 और last digit 5, फिर add करेंगे।

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

    First digit (thousands place) 0 नहीं हो सकता और repetition नहीं हो सकती। First digit के लिए digits 1-9 → 9 choices। First और last digit fix करने के बाद middle दो positions के लिए 8 remaining digits बचते हैं → permutations P(8,2) = 8 × 7 = 56.

    Case A count = 9 × 56 = 504.

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

    Last digit 5 fix है। First digit 0 या 5 नहीं हो सकता → digits {1..9} minus 5 → 8 choices। First और last fix होने के बाद middle positions के लिए फिर 8 digits बचते हैं → P(8,2) = 8 × 7 = 56.

    Case B count = 8 × 56 = 448.

  4. Step 4: Final count (mutually exclusive cases जोड़ें)।

    Total = 504 + 448 = 952.

  5. Final Answer:

    952 four-digit numbers इन conditions को satisfy करते हैं।

  6. Quick Check:

    बिना किसी restriction के no-repetition 4-digit numbers = 9 × 9 × 8 × 7 = 4,536। हमारा answer 952, 4,536 से काफी छोटा है → answer plausible है ✅

Quick Variations

1. Password-type: Repetition allowed हो तो हर digit independent → nr (जैसे 104 = 10,000)।

2. Seating with blocks: साथ बैठने वालों को block मानें और internal permutations से multiply करें।

3. Mixed digit constraints: Divisible-by-2 cases के लिए last-digit even cases में break करें।

4. Multi-stage selection + arrangement: पहले r items चुनें (nCr), फिर उन्हें arrange करें (r!) → same as nPr।

Trick to Always Use

  • Step 1: Problem को independent stages या mutually exclusive cases में break करें।
  • Step 2: हर stage में सही rule चुनें: order हो तो nPr, order ना हो तो nCr, repetition हो तो nr।
  • Step 3: Independent stages → multiply; exclusive cases → add।
  • Step 4: Sanity check: answer हमेशा unconstrained total (जैसे n!, nr, या 2n) से ≤ होना चाहिए।

Summary

Summary

  • हर mixed problem को clear stages या exclusive cases में तोड़ें।
  • Har stage के लिए सही rule तय करें: nPr, nCr या nr।
  • Independent stages को multiply करें और mutually exclusive cases को add करें।
  • Final count को unconstrained possibilities के साथ cross-check करें।

याद रखने वाला example: Constraints को stages में बदलें (जैसे last-digit cases), हर stage में सही rule लगाएँ, और फिर combine करें - इसी तरह example में 952 आया।

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.

Mock Test

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