3.7 Permutation, Combination & Probability in Banking Exams

Permutation, Combination & Probability in Banking Exams

Combinatorics and Probability constitute essential Quantitative Aptitude topics in SBI PO Prelims & Mains. Questions test systematic counting, constrained word arrangements, committee selection, card drawing, and complementary event shortcuts.

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1. Counting Principles, Permutations & Combinations

Fundamental Principles of Counting

• Multiplication Rule (AND): If event A can occur in $m$ ways and event B can occur independently in $n$ ways, both events can occur together in $m \times n$ ways.
• Addition Rule (OR): If event A can occur in $m$ ways and event B can occur in $n$ ways (mutually exclusive), either event can occur in $m + n$ ways.

Permutations vs Combinations Formulas

Useful Combination Properties: $^n C_r = ^n C_{n-r}$, $^n C_0 = 1$, $^n C_n = 1$, $^n C_1 = n$.

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2. Advanced Arrangement Techniques

The String Method (Items Together)

When specific items must ALWAYS sit or stand together, tie them into a single virtual object. Calculate arrangements of virtual objects, then multiply by internal arrangements of tied items.

The Gap Method (Items Never Together)

When specific items must NEVER sit together, arrange the unrestricted items first. Count the available gaps created between unrestricted items, and select gaps for restricted items.

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3. Probability Foundations & Axioms

$$\text{Probability of Event } E = P(E) = \frac{\text{Number of Favorable Outcomes } |E|}{\text{Total Sample Space Outcomes } |S|}$$

• Range: $0 \le P(E) \le 1$.
• Complementary Probability Rule: $$P(E) + P(E') = 1 \implies P(\text{At least one}) = 1 - P(\text{None})$$
• Addition Theorem: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
• Independent Events: Two events A and B are independent if $P(A \cap B) = P(A) \times P(B)$.

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4. Step-by-Step Worked Math Examples

Worked Example 1: Committee Selection with Constraints

Problem: Out of 6 men and 4 women, a committee of 5 people is to be formed. In how many ways can this be done such that the committee contains at least 3 men?

Step-by-Step Solution:

1. "At least 3 men" in a 5-member committee gives 3 mutually exclusive cases:
   • Case 1: 3 Men AND 2 Women
   • Case 2: 4 Men AND 1 Woman
   • Case 3: 5 Men AND 0 Women
2. Compute combinations for each case:
   • Case 1: $^6 C_3 \times ^4 C_2 = \left(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\right) \times \left(\frac{4 \times 3}{2 \times 1}\right) = 20 \times 6 = 120$.
   • Case 2: $^6 C_4 \times ^4 C_1 = ^6 C_2 \times 4 = 15 \times 4 = 60$.
   • Case 3: $^6 C_5 \times ^4 C_0 = 6 \times 1 = 6$.
3. Sum all mutually exclusive cases: Total Ways $= 120 + 60 + 6 = 186 \text{ ways}$.

Worked Example 2: String & Gap Method Word Arrangements

Problem: In how many different ways can the letters of the word 'LEADING' be arranged such that the vowels always come together?

Step-by-Step Solution:

1. Word = LEADING (7 total letters: L, E, A, D, I, N, G).
2. Identify vowels: E, A, I (3 vowels). Consonants: L, D, N, G (4 consonants).
3. Tie the vowels (E, A, I) into 1 single unit string: [EAI].
4. Total items to arrange $= 4 \text{ consonants} + 1 \text{ tied unit} = 5 \text{ items}$.
5. 5 items can be arranged in $5! = 120 \text{ ways}$.
6. The 3 vowels inside the unit can be arranged among themselves in $3! = 6 \text{ ways}$.
7. Total arrangements $= 5! \times 3! = 120 \times 6 = 720 \text{ ways}$.

Worked Example 3: Complementary Probability Shortcut

Problem: An urn contains 5 red, 4 blue, and 3 green balls. If 3 balls are drawn at random simultaneously, what is the probability that at least one red ball is drawn?

Step-by-Step Solution:

1. Total balls $= 5 + 4 + 3 = 12$ balls. Drawing 3 balls gives total sample space $|S| = ^{12} C_3$. $$^{12} C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$
2. Use Complementary Rule: $P(\text{At least 1 Red}) = 1 - P(\text{No Red Ball})$.
3. "No Red Ball" means all 3 balls are drawn from the remaining non-red balls (4 blue + 3 green $= 7$ non-red balls).
4. Favorable outcomes for zero red balls $= ^7 C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
5. $P(\text{No Red Ball}) = \frac{35}{220} = \frac{7}{44}$.
6. $P(\text{At least 1 Red}) = 1 - \frac{7}{44} = \frac{37}{44}$.

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5. Common Pitfalls in Banking Probability

• Confusing Permutations and Combinations: Always check if order creates distinct outcomes. For example, selecting 2 captains from 10 players requires $^n C_r$ if roles are identical, but $^n P_r$ if one is Captain and one is Vice-Captain.
• Without Replacement Probability: When drawing items sequentially without replacement, adjust sample space size for subsequent draws ($12 \to 11 \to 10$).