How to Solve: CUET Quant – Permutation, Combination & Probability

How to Solve: CUET Quant – Permutation, Combination & Probability

Introduction

"Imagine you’re picking a 3-person team from 10 friends—how many ways can you do it? Or, if you draw 2 cards from a deck, what’s the chance both are aces? These aren’t just party tricks—they’re CUET exam questions worth 5-10 marks. Master this, and you’ll solve them in under 2 minutes."

What You Need To Know First

1. Factorials (n!) – The product of all positive integers up to n (e.g., 4! = 4 × 3 × 2 × 1 = 24).
2. Basic Counting Principle – If Event A has m ways and Event B has n ways, the total ways for both is m × n.
3. Difference Between "And" vs. "Or" – "And" means multiply probabilities; "Or" means add (but subtract overlap if events aren’t mutually exclusive).

Key Vocabulary

Formulas To Know

1. Permutation (Order Matters)

Formula: [ nP_r = \frac{n!}{(n - r)!} ] - n = Total items. - r = Items to arrange. - MEMORISE THIS (Not given on CUET sheet).

When to use: Arranging books on a shelf, forming passwords, ranking teams.

2. Combination (Order Doesn’t Matter)

Formula: [ nC_r = \frac{n!}{r!(n - r)!} ] - n = Total items. - r = Items to choose. - MEMORISE THIS (Not given on CUET sheet).

When to use: Selecting a committee, choosing lottery numbers, forming groups.

3. Probability of an Event

Formula: [ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} ] - MEMORISE THIS (Given on exam sheet, but know how to apply it).

When to use: Calculating chances of drawing a card, rolling a number, or passing a test.

4. Probability of Independent Events (A and B)

Formula: [ P(A \text{ and } B) = P(A) \times P(B) ] - MEMORISE THIS (Given on exam sheet).

When to use: Flipping two coins, drawing two cards with replacement.

5. Probability of Mutually Exclusive Events (A or B)

Formula: [ P(A \text{ or } B) = P(A) + P(B) ] - MEMORISE THIS (Given on exam sheet).

When to use: Rolling a 2 or a 5 on a die.

6. Probability of Non-Mutually Exclusive Events (A or B)

Formula: [ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) ] - MEMORISE THIS (Given on exam sheet).

When to use: Drawing a king or a heart from a deck (king of hearts is counted twice if you don’t subtract).

Step-by-Step Method

Step 1: Read the Question Carefully

• Underline keywords: "arrange," "select," "probability," "with/without replacement," "order matters."
• Circle numbers: total items (n), items to choose/arrange (r).

Step 2: Decide if Order Matters

• Order matters? → Use Permutation (nPr).
• Order doesn’t matter? → Use Combination (nCr).

Step 3: Write Down the Formula

• Plug in n and r into the correct formula.
• If probability, identify favorable and total outcomes.

Step 4: Calculate Factorials

• Simplify before multiplying (e.g., 5! / 3! = 5 × 4).
• Use calculator for large numbers (e.g., 10! / 7! = 10 × 9 × 8).

Step 5: Solve for Probability (If Needed)

• For "and" → Multiply probabilities.
• For "or" → Add probabilities (and subtract overlap if needed).

Step 6: Check Units and Final Answer

• Does the answer make sense? (e.g., probability should be between 0 and 1).
• Write the answer in simplest form (e.g., 1/2 instead of 0.5).

Worked Example Using Steps

Question: How many ways can you arrange 3 books out of 5 on a shelf?

Step 1: Keywords: "arrange" (order matters), 3 books out of 5 → n = 5, r = 3. Step 2: Order matters → Use Permutation (nPr). Step 3: Formula: ( 5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} ). Step 4: Calculate: ( \frac{5 × 4 × 3 × 2!}{2!} = 5 × 4 × 3 = 60 ). Step 5: Not a probability question → Skip. Step 6: Final answer: 60 ways.

Worked Examples

Example 1 – Basic (Combination)

Question: In how many ways can you choose 2 students from a group of 6?

Solution: 1. Keywords: "choose" (order doesn’t matter), 2 from 6 → n = 6, r = 2. 2. Order doesn’t matter → Use Combination (nCr). 3. Formula: ( 6C_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} ). 4. Simplify: ( \frac{6 × 5 × 4!}{2 × 1 × 4!} = \frac{30}{2} = 15 ). 5. Final answer: 15 ways.

What we did and why: We used combination because the order of selection doesn’t matter (choosing Alice and Bob is the same as Bob and Alice).

Example 2 – Medium (Probability with Replacement)

Question: A bag has 4 red and 3 blue balls. If you draw 2 balls with replacement, what’s the probability both are red?

Solution: 1. Keywords: "probability," "with replacement" (independent events). 2. Total balls = 4 red + 3 blue = 7. 3. Probability of first red: ( P(\text{Red}_1) = \frac{4}{7} ). 4. Probability of second red (with replacement): ( P(\text{Red}_2) = \frac{4}{7} ). 5. Both red: ( P(\text{Red}_1 \text{ and Red}_2) = \frac{4}{7} × \frac{4}{7} = \frac{16}{49} ). 6. Final answer: 16/49.

What we did and why: Since the ball is replaced, the two draws are independent, so we multiply probabilities.

Example 3 – Exam Style (Disguised Permutation)

Question: A password consists of 3 distinct letters from A, B, C, D, E. How many possible passwords are there?

Solution: 1. Keywords: "password," "distinct letters" (order matters, no repeats). 2. Total letters (n) = 5, letters to choose (r) = 3. 3. Order matters → Use Permutation (nPr). 4. Formula: ( 5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} ). 5. Calculate: ( \frac{5 × 4 × 3 × 2!}{2!} = 5 × 4 × 3 = 60 ). 6. Final answer: 60 passwords.

What we did and why: The word "password" implies order matters (ABC ≠ BAC), so we used permutation.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in. Permutations are for order matters—like arranging books or passwords. Use ( nP_r = \frac{n!}{(n-r)!} ). Combinations are for order doesn’t matter—like picking teams or toppings. Use ( nC_r = \frac{n!}{r!(n-r)!} ). Probability is just favorable over total. For ‘and,’ multiply; for ‘or,’ add—but subtract the overlap if they can happen together. Watch out for traps: ‘at least one’ means 1 minus the chance of none. And always check if the question is about selection or arrangement. You’ve got this—go crush those CUET questions!