In this guide
Permutation–combination (P&C) and probability do not carry many questions in the SBI PO paper, but they turn up often enough in the prelims and as one side of a quantity comparison that skipping them is a poor trade. The good news is that the whole topic rests on two counting rules and one definition. Learn those properly and most questions take under a minute.
The risk is the opposite of most quant topics. The arithmetic is easy; the trap is counting the wrong thing. So this guide spends as much time on "what am I counting?" as on formulas.
The two counting rules
- AND means multiply. If a task has two steps, with m ways for the first and n ways for the second, the whole task can be done in m × n ways.
- OR means add. If a task can be done in one of two separate ways, with m and n options, there are m + n ways in all.
Why. Every choice in the first step can be paired with every choice in the second, which is why the counts multiply. Separate routes do not overlap, so they simply add.
Arrange or choose?
| Question wording | Order matters? | Use |
|---|---|---|
| Arrange, seat, line up, form a word or a number | Yes | Permutation: nPr = n! ÷ (n − r)! |
| Choose, select, form a committee or a team, pick a hand | No | Combination: nCr = n! ÷ (r! × (n − r)!) |
Why nCr divides by r!. Every group of r people can be lined up in r! orders. A permutation counts each of those orders separately; a combination counts the group once. So nCr = nPr ÷ r!.
Two shortcuts save time: nCr = nC(n − r), so 9C7 = 9C2 = 36; and nC1 = n.
Arrangement rules
| Situation | Count | Why |
|---|---|---|
| n different items in a row | n! | n choices for the first place, n − 1 for the next, and so on |
| n items with p alike and q alike | n! ÷ (p! × q!) | Swapping identical letters gives the same word, so divide out those swaps |
| n people round a table | (n − 1)! | Rotating everyone one seat gives the same arrangement, so fix one person |
| Some items must be together | Treat them as one block, then multiply by the block's internal arrangements | The block moves as one unit |
| Some items must never be together | Total − (arrangements with them together) | Easier to count the opposite |
Worked examples: counting
Example 1. How many arrangements are there of the letters of BANK?
- Four different letters: 4! = 24.
Example 2. Of APPLE?
- Five letters with P twice: 5! ÷ 2! = 120 ÷ 2 = 60.
Example 3. Vowels together. In how many ways can the letters of ORANGE be arranged so that the vowels are always together?
- Vowels O, A, E form one block. With R, N and G that makes 4 units: 4! = 24.
- Inside the block the three vowels can be arranged in 3! = 6 ways.
- 24 × 6 = 144.
Example 4. Committee. A team of 2 officers from 5 and 2 clerks from 4 is to be formed. How many ways?
- Officers AND clerks, so multiply: 5C2 × 4C2 = 10 × 6 = 60.
Example 5. "At least one". From 4 officers and 3 clerks, a committee of 3 is chosen. In how many ways does it include at least one clerk?
- All committees: 7C3 = 35. Committees with no clerk: 4C3 = 4.
- 35 − 4 = 31.
Example 6. Digits. How many 3-digit numbers can be formed from 1, 2, 3, 4 and 5 without repeating a digit? How many of them are even?
- All: 5 × 4 × 3 = 60.
- Even: the last digit must be 2 or 4 (2 ways). Then 4 choices for the first digit and 3 for the middle: 2 × 4 × 3 = 24.
- Fill the restricted place first. That is the habit that prevents most digit errors.
Example 7. In how many ways can six people sit round a table?
- (6 − 1)! = 5! = 120.
Probability: one definition and three rules
Probability = favourable outcomes ÷ total outcomes, when every outcome is equally likely.
- Complement: P(not A) = 1 − P(A). Use it for "at least one".
- AND for independent events: multiply the probabilities.
- OR for events that cannot happen together: add. If they can overlap, subtract the overlap once.
Facts to have ready: a coin has 2 outcomes; n coins have 2ⁿ; one die has 6; two dice have 36. A pack has 52 cards in four suits of 13; 26 red and 26 black; 4 of each rank; 12 face cards (jack, queen and king in each suit).
Worked examples: probability
Example 8. Same colour. A bag has 5 red and 4 green balls. Two are drawn together. What is the probability that both are the same colour?
- Both red: 5C2 = 10. Both green: 4C2 = 6. All pairs: 9C2 = 36.
- (10 + 6) ÷ 36 = 4/9.
Example 9. Overlap. One card is drawn from a pack of 52. What is the probability that it is red or a king?
- 26 red cards + 4 kings − 2 red kings counted twice = 28.
- 28 ÷ 52 = 7/13.
Example 10. Coins. Three coins are tossed. What is the probability of exactly two heads?
- HHT, HTH and THH: 3 of 8 outcomes. 3/8.
Example 11. Dice sum. Two dice are thrown. What is the probability that the sum is 8?
- (2, 6), (3, 5), (4, 4), (5, 3), (6, 2): 5 of 36. 5/36.
Example 12. At least one. Two dice are thrown. What is the probability of at least one six?
- No six on either die: 5 × 5 = 25 of 36. So at least one six: 1 − 25/36 = 11/36.
- Counting directly gives the same: six on the first (6) + six on the second (6) − both (1) = 11.
Example 13. One after another. A bag has 3 red and 2 blue balls. Two are drawn one after the other without replacement. What is the probability that both are red?
- First red: 3/5. Second red, given the first was red: 2/4.
- 3/5 × 2/4 = 6/20 = 3/10.
- Check with combinations: 3C2 ÷ 5C2 = 3 ÷ 10. "Drawn together" and "drawn one by one without replacement" give the same answer.
Common mistakes
- Missing the overlap in OR questions. "Red or king" counts the two red kings once, not twice.
- Forgetting repeated letters. LETTER has two Es and two Ts; divide by 2! twice.
- Counting "at least one" directly when the complement is one line.
- Replacing when the question says without replacement, or the reverse. With replacement, the second draw has the same total as the first.
- Assuming (n − 1)! for a row. It applies only to circles.
Practice set
- How many arrangements are there of the letters of LETTER?
- In how many ways can 4 people be chosen from 9?
- Two dice are thrown. What is the probability of a doublet (both showing the same number)?
- A bag has 3 white and 5 black balls. Two are drawn together. What is the probability of one of each colour?
- In how many ways can 5 different books be arranged on a shelf if two particular books must be together?
- One card is drawn from a pack of 52. What is the probability that it is a face card?
- Two dice are thrown. What is the probability that the sum is a prime number?
- A committee of 4 is chosen at random from 5 officers and 4 clerks. What is the probability that it has exactly 2 clerks?
Answers:
- 180. 6! ÷ (2! × 2!) = 720 ÷ 4.
- 126. 9C4 = (9 × 8 × 7 × 6) ÷ 24.
- 1/6. Six doublets out of 36.
- 15/28. 3 × 5 = 15 mixed pairs out of 8C2 = 28.
- 48. The pair is one block: 4! = 24 arrangements, × 2 for the order inside the block.
- 3/13. 12 face cards out of 52.
- 5/12. Sums 2, 3, 5, 7 and 11 occur 1, 2, 4, 6 and 2 times: 15 of 36.
- 10/21. Favourable 5C2 × 4C2 = 60; total 9C4 = 126; 60/126 = 10/21.
What to do next
- Before solving any P&C question, write "arrange" or "choose" next to it.
- Learn the card and dice facts above until you do not need to count them.
- Practise ten "at least one" questions using the complement only.
- Use P&C values as quantities in quantity comparison, and keep the formulas on your quant revision sheet.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the State Bank of India website .
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