Permutation and combination for IBPS PO

Permutation and combination is counting without listing. The whole topic rests on one question you must ask before you touch a formula: does the order matter? If swapping two items gives a new outcome, you are arranging. If it does not, you are choosing.

In IBPS PO, the topic appears as standalone questions in the prelims and, more often, underneath probability, where every bag-of-balls question is a combination count. The level stays moderate: letters, committees, digits and round tables. Learn the reasoning behind each formula, and the variations stop being surprises.

Arrange or choose?

The formulas and why they work

• Counting principle. If one step can be done in m ways and the next in n ways, both together can be done in m × n ways. Every formula below comes from this.
• n different things in a row: n! The first place has n choices, the next n − 1, and so on.
• Repeated items: n! ÷ (p! × q! …) for items repeated p, q … times. Why: swapping two identical Ps gives the same word, so every distinct word was counted p! times.
• Choosing r from n: nCr = n! ÷ (r! × (n − r)!). Why: arrange r items from n (that is nPr), then divide by r!, because a chosen group can be ordered in r! ways and they all count as one group.
• Circular: (n − 1)! Why: around a table, turning everyone one seat to the left gives the same arrangement. Fix one person's seat and arrange the rest.
• Handy results: handshakes among n people = nC2; diagonals of an n-sided polygon = n(n − 3)/2; nCr = nC(n − r).

Worked examples

Example 1 (repeated letters): In how many ways can the letters of APPLE be arranged?

• 5 letters with P twice: 5! ÷ 2! = 120 ÷ 2 = 60.

Example 2 (vowels together and apart): The word ORANGE has vowels O, A, E and consonants R, N, G.

• All vowels together: glue O, A, E into one block. Now arrange 4 units (block, R, N, G): 4! = 24. Arrange inside the block: 3! = 6. Total = 144.
• Vowels not all together: total 6! = 720, minus 144 = 576.
• No two vowels together: arrange the consonants: 3! = 6. They create 4 gaps: before R, between R and N, between N and G, and after G. Place the 3 vowels in 3 of the 4 gaps in order: 4 × 3 × 2 = 24. Total = 6 × 24 = 144.
• The last two answers differ because "not all together" still allows two vowels side by side.

Example 3 (committee with "at least"): A committee of 4 is to be formed from 5 men and 4 women, with at least 2 women. In how many ways?

• 2 women and 2 men: 4C2 × 5C2 = 6 × 10 = 60.
• 3 women and 1 man: 4C3 × 5C1 = 4 × 5 = 20.
• 4 women: 4C4 = 1.
• Total = 81.
• Check with the complement: all committees 9C4 = 126, minus 0 women (5C4 = 5) and 1 woman (4 × 5C3 = 40), gives 126 − 45 = 81.

Example 4 (numbers with 0): Using the digits 0, 1, 2, 3, 4 and 5 without repetition, how many 4-digit numbers can be formed? How many of them are even?

• The first digit cannot be 0: 5 choices. Then 5, 4 and 3 choices. Total = 5 × 5 × 4 × 3 = 300.
• Even, ending in 0: the other three places take 5 × 4 × 3 = 60.
• Even, ending in 2 or 4: 2 choices for the last digit. The first digit cannot be 0 or the last digit: 4 choices. The middle two: 4 × 3. Total = 2 × 4 × 12 = 96.
• Even numbers = 60 + 96 = 156.

Example 5 (circular with a condition): In how many ways can 6 people sit around a round table if two particular people must sit together?

• Glue the two into one unit: 5 units around a table give (5 − 1)! = 24.
• The pair can swap: × 2. Total = 48.

Example 6 (include one, exclude one): A team of 5 is to be chosen from 8 players. One particular player must be in the team, and another particular player must be left out. In how many ways?

• One place is filled and one player is removed, so choose 4 from the remaining 6: 6C4 = 15.

Common mistakes

• Using combination when order matters, or the reverse. Ask "does swapping create a new outcome?" before anything else.
• Forgetting to divide for repeated letters. LEVEL has two Ls and two Es.
• Forgetting the internal arrangement of a glued block.
• Confusing "not all together" with "no two together".
• Letting 0 lead a number.
• Using n! for a round table instead of (n − 1)!.
• Adding when you should multiply. "And" (both steps happen) multiplies; "or" (separate cases) adds, as in Example 3.

Practice set

1. In how many ways can the letters of LEVEL be arranged?
2. Find 8C3.
3. In how many ways can 6 people sit around a round table?
4. A team of 3 is chosen from 6 boys and 4 girls with exactly one girl. In how many ways?
5. How many 3-digit numbers can be formed from 1, 2, 3, 4 and 5 without repeating a digit?
6. In how many ways can the letters of BANKER be arranged so that the vowels are together?
7. Twelve people are at a meeting, and each shakes hands once with every other. How many handshakes are there?
8. A committee of 3 is chosen from 6 men and 5 women. In how many ways can it include at least one woman?

Answers:

1. 30. 5! ÷ (2! × 2!) = 120 ÷ 4.
2. 56. (8 × 7 × 6) ÷ 6.
3. 120. (6 − 1)! = 5!.
4. 60. 4C1 × 6C2 = 4 × 15.
5. 60. 5 × 4 × 3.
6. 240. Glue A and E: 5 units give 5! = 120, and the block gives 2! = 2.
7. 66. 12C2 = (12 × 11) ÷ 2.
8. 145. All committees 11C3 = 165, minus all-men 6C3 = 20.

What to do next

• For a week, write "order matters: yes/no" before solving any counting question. The habit fixes most errors.
• Redo Examples 2 and 4. Vowel conditions and digits with 0 are the variations that trip people up most often.
• Move straight to probability and solve five bag questions using the combination counts from here.