Permutation and Combination for SBI PO — Complete Strategy Guide

Concept

At its core, this chapter is about one decision: does order matter?

If you are choosing 3 students from a class of 10 to sit in a row for a photograph, the position of each person changes the arrangement — that is a permutation (क्रमचय). If you are choosing 3 students to form a committee, a group is the same group regardless of who was picked first — that is a combination (संचय).

Here is the analogy that sticks: think of a lock versus a lottery. A combination lock is actually misnamed — it should be called a permutation lock, because 3-1-5 and 5-1-3 open different locks. A lottery, where you just need the right 5 numbers regardless of draw order, is the true combination scenario.

The two formulas flow directly from this:

$^nP_r = \frac{n!}{(n-r)!}$

$^nC_r = \frac{n!}{r!(n-r)!}$

The relationship between them is equally important:

$^nP_r = ^nC_r \times r!$

This makes sense: once you have selected $r$ items (combination), multiplying by $r!$ accounts for all the ways those $r$ items can be ordered among themselves.

Why does SBI PO push this topic to advanced difficulty? Because the exam does not test bare formula substitution. It tests three higher-order applications — repeated letters in word arrangements, the gap method for non-adjacency constraints, and stars-and-bars for identical-object distribution. These three patterns account for the bulk of PYQs. Every concept in this page is aimed at cracking exactly those three patterns.

One more grounding point: $0! = 1$ by definition, and $^nC_0 = ^nC_n = 1$. Do not second-guess these under exam pressure.

Deep Dive

Factorial and the Counting Principle

The fundamental counting principle states: if event A can happen in $m$ ways and event B in $n$ ways, both together happen in $m \times n$ ways. Factorial $n!$ is simply this principle applied to arranging $n$ distinct objects: $n \times (n-1) \times (n-2) \times \cdots \times 1$.

Quick recall: $5! = 120$, $6! = 720$, $7! = 5040$, $8! = 40320$, $9! = 362880$, $10! = 3628800$. Memorise up to $10!$ — you will use these as lookup values, not compute them from scratch.

Permutations with Repetition

When a word has repeated letters, the formula becomes:

$\text{Arrangements} = \frac{n!}{p! \cdot q! \cdot r! \cdots}$

where $p, q, r$ are frequencies of each repeated letter.

Example: PUNCTUAL has 8 letters, with U appearing twice. Answer = $\frac{8!}{2!} = \frac{40320}{2} = 20160$.

The logic: without the denominator, you count every swap of the two U's as a new arrangement, which is wrong since both U's are identical.

The Gap Method (Non-Adjacency Constraint)

This is the most tested technique. When a problem says "no two [group X] should be together," you:

1. First arrange the other group (say $m$ people) in $m!$ ways. This creates $m+1$ gaps — one before the first person, one between each consecutive pair, and one after the last.
2. Choose $r$ gaps from those $m+1$ gaps for group X: $^{m+1}C_r$ ways.
3. Arrange group X within chosen gaps: $r!$ ways.
4. Multiply: $m! \times {^{m+1}C_r} \times r!$

For 5 boys and 4 girls, no two girls together:

• Arrange 5 boys: $5! = 120$ ways
• Gaps created: $5 + 1 = 6$ gaps
• Choose 4 of 6 gaps for girls: $^6C_4 = 15$
• Arrange 4 girls: $4! = 24$
• Total: $120 \times 15 \times 24 = 43200$

Treating a Group as One Unit (Always-Together Constraint)

When a problem says a specific set of people must always be together:

1. Bind those people into one super-unit.
2. Arrange the reduced total (original $n$ minus group size + 1) in that many factorial ways.
3. Internally arrange the bound group: (group size)! ways.
4. Multiply.

For 8 people with 3 always together: treat 3 as one unit → 6 units → $6! \times 3! = 720 \times 6 = 4320$.

Complementary Counting

When "at least one" of something is required, it is almost always faster to subtract the unwanted case from the total:

$\text{At least one} = \text{Total} - \text{None at all}$

From 6 novels + 3 poems, choose 4 with at least 1 poem:

• Total: $^9C_4 = 126$
• No poems (all novels): $^6C_4 = 15$
• Answer: $126 - 15 = 111$

Train yourself to reach for complementary counting the moment you see "at least one."

Stars and Bars (Identical Object Distribution)

This pattern appears repeatedly in SBI PO under distribution problems. The formula for distributing $n$ identical objects into $k$ distinct groups with no restriction (groups can receive 0):

$\text{Ways} = {^{n+k-1}C_{k-1}}$

When each group must receive at least 1, first give 1 to each group, reducing $n$ to $n-k$, then apply the unrestricted formula:

$\text{Ways} = {^{(n-k)+(k-1)}C_{k-1}} = {^{n-1}C_{k-1}}$

For 10 identical books to 4 students, each getting at least 1:

• After giving 1 each, 6 books remain
• $^{6+4-1}C_{4-1} = {^9C_3} = 84$

Note that $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$.

Key nCr Values to Recall Instantly

| Expression | Value | |---|---| | $^9C_3$ | 84 | | $^9C_4$ | 126 | | $^6C_4$ | 15 | | $^7C_4$ | 35 | | $^8C_3$ | 56 |

These come up directly in PYQs. Compute them once; do not recompute under time pressure.

Memory Tricks & Shortcuts

Fast-Solving Framework

Read the problem. Then follow this decision path:

Step 1 — Identical or distinct objects?

• If identical → Stars and Bars territory.
• If distinct → continue.

Step 2 — Does order matter?

• Seating in a row, forming queues, filling designated posts = YES → Permutation ($^nP_r$).
• Forming groups, committees, teams, selecting items = NO → Combination ($^nC_r$).

Step 3 — Is there a constraint?

• "Always together" → Bind into one unit, reduce count, multiply by internal arrangements.
• "Never together / no two adjacent" → Gap method.
• "At least one of type X" → Total minus none-of-type-X.
• "Each group gets at least one" → Stars and bars with $n-k$ remaining objects.

Step 4 — Repeated elements?

• Word problems → check for repeated letters → divide by their factorials.

Step 5 — Compute using memorised factorial/nCr values.

If a problem seems to combine two constraints, apply them one at a time in the order: arrangement first, then constraint second. Never try to build both simultaneously.

Solved PYQs

Why this question: Tests the most fundamental word-arrangement formula with a repeated-letter twist that catches unprepared candidates.

Solving path: Write PUNCTUAL, count 8 letters. Scan for repeats: U appears at positions 2 and 7 — twice. No other repeats. Apply formula: $\frac{8!}{2!} = \frac{40320}{2} = 20160$. The trap option 40320 (which is $8!$) is there for candidates who miss the repeated U. The option 960 would require a very specific additional constraint that does not exist — eliminate it immediately.

Why this question: Tests the gap method under a non-adjacency constraint — one of the three core SBI PO P&C patterns.

Solving path: Draw the 5 boy slots mentally: _ B _ B _ B _ B _ B _ → 6 gaps. Arrange 5 boys: $5! = 120$. Choose 4 of 6 gaps for girls: $^6C_4 = 15$. Arrange 4 girls: $4! = 24$. Multiply: $120 \times 15 \times 24 = 120 \times 360 = 43200$.

Why this question: Tests the complementary counting reflex — the fastest path to "at least one" problems.

Solving path: Total books = 6 + 3 = 9. Choose 4: $^9C_4 = 126$. All 4 from novels only: $^6C_4 = 15$. At least 1 poem = $126 - 15 = 111$. Do not enumerate by exact poem count — the subtraction route is 4 steps versus 9.

Why this question: Tests stars and bars — the pattern that separates 80-percentile scorers from 95-percentile scorers on this topic.

Solving path: Identical books = 10, students = 4, each gets at least 1. Give 1 to each student: 6 books remain. Now distribute 6 identical books among 4 students freely: $^{6+4-1}C_{4-1} = {^9C_3} = \frac{9 \times 8 \times 7}{6} = 84$. Options 70, 126, 120 are planted for common formula errors ($^8C_3$, $^9C_4$, $5!$).

Why this question: Tests the always-together binding technique with a larger group, checking whether you correctly compute 6 units rather than 8.

Solving path: Bind the 3 particular people into one unit → total units = 8 − 3 + 1 = 6. Arrange 6 units: $6! = 720$. Arrange 3 people within their unit: $3! = 6$. Total = $720 \times 6 = 4320$. The trap answer 5040 is $7!$ — for candidates who reduce to 7 units instead of 6.

Common Mistakes

• Missing repeated letters in word problems. Always scan the full word before writing $n!$. PUNCTUAL, BALLOON, MATHEMATICS, MISSISSIPPI are classic traps. A missed repeat will land you on the wrong answer every time.

• Using $^mC_r$ instead of $^{m+1}C_r$ in the gap method. The gaps are always one more than the number of people in the fixed group. Five boys create six gaps, not five. Draw the visual if unsure.

• Forgetting to multiply by $r!$ when the restricted group has internal arrangements. $^6C_4$ gives you the chosen gaps, but girls can be arranged within those gaps in $4!$ ways. Leaving out $4!$ gives 1800 instead of 43200.

• Applying stars and bars to distinct objects. Stars and bars applies only when objects are identical (or interchangeable). Distributing 10 distinct books among 4 students is a completely different problem ($4^{10}$ without restriction, not $^{13}C_3$).

• Using Total − None when the complement is more complex than a single case. Complementary counting is fastest for "at least one of one type." For "at most two from each group," direct enumeration or inclusion-exclusion may be necessary. Do not force the subtraction method on every constraint.

• Confusing $^{n-1}C_{k-1}$ with $^{n+k-1}C_{k-1}$. When the problem says each box/student must get at least 1, the formula simplifies to $^{n-1}C_{k-1}$ directly. When there is no such restriction, use $^{n+k-1}C_{k-1}$. Mixing these two is the most common stars-and-bars error and it produces a plausible-looking wrong answer that is usually present as a distractor option.

Related Topics

• probability basics
• probability advanced
• number series
• data interpretation
• simplification
• quadratic equations

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