PROJECTENGRAM

COUNTING BY CASES

Tags: #◇Computer Science #◇STT101A #◇Conceptual #◇Counting by Cases


◇Overview

Counting by Cases ( the Addition Principle / Rule of Sum) decomposes a complex event EE into kk mutually exclusive and collectively exhaustive subcases E1,E2,…,EkE_1, E_2, \dots, E_k. Within each case, the Multiplication Rule counts the configurations; across disjoint cases, the counts are summed:
∣E∣=∣E1∣+∣E2∣+⋯+∣Ek∣|E| = |E_1| + |E_2| + \cdots + |E_k|

◇Key Definitions

  • Addition Principle (Rule of Sum): If E1,E2,…,EkE_1, E_2, \dots, E_k are pairwise disjoint (Ei∩Ej=∅E_i \cap E_j = \emptyset for i≠ji \ne j), then ∣⋃i=1kEi∣=∑i=1k∣Ei∣\left|\bigcup_{i=1}^k E_i\right| = \sum_{i=1}^k |E_i|.
  • Case-Splitting Strategy: Used whenever constraints change depending on an earlier choice (e.g., "at least kk", "at most kk", or digit restrictions where 00 cannot be a leading digit but can be the last digit of an even number).

◇Worked Example

Problem: From 66 men and 55 women, a 44-person committee must have at least 33 women. How many such committees can be formed?

Solution:
Split into two mutually exclusive cases based on the exact number of women:

  • Case 1 (Exactly 3 women and 1 man): (53)×(61)=10×6=60\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60.
  • Case 2 (Exactly 4 women and 0 men): (54)×(60)=5×1=5\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5.
  • Total: 60+5=6560 + 5 = 65.

◇References & Navigation

CONFIRMATION // 理解度の確認

ACTIVE RECALL VERIFICATION

Select an answer to verify conceptual comprehension and reveal the technical rationale.

Q1

What condition MUST be satisfied by the cases E1,E2,…,EkE_1, E_2, \dots, E_k when applying Counting by Cases (∣E∣=∑∣Ei∣|E| = \sum |E_i|)?

Q2

Why is the following attempt to count 4-person committees with at least 1 woman from 55 women and 66 men INCORRECT: (51)×(103)=5×120=600\binom{5}{1} \times \binom{10}{3} = 5 \times 120 = 600?

Q3

Using valid Counting by Cases (or the complement), how many 44-person committees from 55 women and 66 men have at least 11 woman?

Q4

How many 33-digit even integers with distinct digits can be formed from {0,1,2,3,4,5}\{0, 1, 2, 3, 4, 5\} (where the hundreds digit cannot be 00)?

Q5

Why must the problem in Question 4 be split into Case 1 (units digit is 00) and Case 2 (units digit is 22 or 44)?