How do you calculate nCr and nCr? This simple guide explains permutations and combinations in easy words. Imagine setting a 4-digit PIN: every time you swap two digits, you get a completely different PIN, so the order of digits matters hugely.But imagine picking 3 pizza toppings from 10: the order doesn't change what will end up on your pizza. That's the difference. Mathematicians needed two separate counting tools to handle these two very different situations. nPr (permutations) counts arrangements where order matters. nCr (combinations) counts selections where order doesn't. Learn the formulas, step-by-step calculations, and practice problems to master counting.

nCr and nPr Formulas: Explained
How to Calculate nCr and nPr: Step-by-Step
How to Decide: Use nPr or nCr?
Difference Between nPr and nCr
Frequently Asked Questions of nCr and nPr

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nCr and nPr Formulas: Explained

${}^{n}\mathrm{P}_{r}$ : Permutation Formula

${}^{n}\mathrm{P}_{r} = \frac{n! }{(n − r)!}$

Where:

n = total number of items available
r = number of items you are arranging
condition: r ≤ n

${}^{n}\mathrm{C}_{r}$ : Combination Formula

${}^{n}\mathrm{C}_{r}= \frac{n! }{[r! × (n − r)!]}$

Where:

n = total number of items available
r = number of items you are selecting
condition: r ≤ n

Relation between nCr and nPr

The relationship between nCr (combination) and nPr (permutation) helps us understand how choosing and arranging objects are connected.
${}^{n}\mathrm{C}_{r}=\frac{ {}^{n}\mathrm{P}_{r} }{r!}$

This means that after choosing r objects from n objects using nCr, we can arrange those selected objects in r! different ways to get nPr.

For example, choosing 3 students from a group of 10 for a committee is a combination because the order does not matter: 10C3. However, selecting a president, vice-president and secretary from the same group is a permutation because each position creates a different arrangement: 10P3

How to Calculate nCr and nPr: Step-by-Step

How to Calculate nPr:

Step 1: Identify the values of n and r
Find the total number of objects (n) and the number of objects to be arranged (r).

Step 2: Write the nPr formula

nPr = n! / (n-r)!

Step 3: Substitute the values of n and r into the formula

Step 4: Calculate the factorial values
Expand the factorials and simplify.

Step 5: Divide the numerator by the denominator
The final answer is the number of possible arrangements.

Example 1: You are organising a 100m sprint race with 5 participants. How many different ways can the gold, silver, and bronze medals be awarded?

Solution: Given: n = 5 (total runners), r = 3 (medals to award)

Here the order matters as gold ≠ silver ≠ bronze

Write the nPr formula and substitute:

5P3 = ${}^{5}\mathrm{P}_{3}$ = 5! / (5 − 3)! = 5! / 2!

Expand 5! and 2!:

5! = 5 × 4 × 3 × 2 × 1 = 120

2! = 2 × 1 = 2

Divide:

${}^{5}\mathrm{P}_{3}$ = 120 / 2 = 60

So 5P3 = 60 different medal arrangements are possible.

How to Calculate nCr:

Step 1: Identify the values of n and r
Find the total number of objects (n) and the number of objects to be selected (r).

Step 2: Write the nCr formula

nCr = n! / r!(n-r)!

Step 3: Substitute the values of n and r into the formula

Step 4: Calculate the factorial values
Expand the factorials and simplify.

Step 5: Divide the numerator by the denominator
The final answer gives the number of possible selections.

Example 2: A coach must select 3 players from a squad of 5 for a penalty shootout team. How many different teams of 3 can be formed?

Solution: Given: n = 5 (total players), r = 3 (players to select)

Here the order doesn't matter as team {A, B, C} = {C, A, B}

Write the nCr formula and substitute:

5C3 = ${}^{5}\mathrm{C}_{3}$ = 5! / [3! × (5 − 3)!] = 5! / (3! × 2!)

Expand all factorials:

5! = 120, 3! = 6 and 2! = 2

Compute denominator, then divide:

${}^{5}\mathrm{C}_{3}$ = 120 / (6 × 2) = 120 / 12 = 10

Hence, ${}^{5}\mathrm{C}_{3}$ = 10 different teams of 3 can be formed.

How to Decide: Use nPr or nCr?

It is important to correctly identify which formula to use.

Ask yourself: "If I swap two of my chosen items, do I get a different answer?"

YES → Use nPr	NO → Use nCr
→ Awarding 1st, 2nd, 3rd place	→ Selecting a sports team
→ Creating passwords or PINs	→ Choosing pizza toppings
→ Assigning different job roles	→ Forming a committee
→ Arranging books on a shelf	→ Picking lottery numbers
→ Seating people in specific seats	→ Choosing questions from a paper
→ Scheduling tasks in order	→ Distributing identical items

Difference Between nPr and nCr

Feature	nPr (Permutation)	nCr (Combination)
Full Name	Permutation of n, r	Combination of n, r
Formula	$\dfrac{n!}{(n-r)!}$	$\dfrac{n!}{r!(n-r)!}$ }
Order	Matters: {A, B} ≠ {B, A}	Irrelevant: {A, B} = {B, A}
Value	Always ≥ nCr (for the same n, r)	Always ≤ nPr (for the same n, r)
Keyword Clues	"arrange", "rank", "assign", "order", "schedule", "first/second/third"	"choose", "select", "pick", "group", "team", "committee"
Also Written As	P(n,r) or $^nP_r$	C(n,r) or $^nC_r$
Special Values	nP0 = 1, nPn = n!	nC0 = 1, nCn = 1, nC1 = n

Know more about related topics:

Frequently Asked Questions of nCr and nPr

1. What is the differnce between nPr and nCr?

The difference is whether order matters. nPr (permutation) counts the number of ways to arrange r items from n, where every different ordering is a different result. nCr counts the number of ways to choose r items from n, where only the group matters and ordering is ignored.

2. Why does nCr formula have r! in the denominator but nPr dosent?

For any group of r items, there are r! ways to arrange them. So when you compute nPr, you're already counting those r! orderings per group. Since nCr doesn't care about order, it divides nPr by r!

3. Is nCr same as nC(n−r)?

Yes. This is the symmetry property of combinations: nCr = nC(n−r). For example, 10C4 = 10C6 = 210.

4. Can r be greater than n in nCr or nPr?

No. Both nCr and nPr are only defined when r ≤ n. You cannot choose or arrange more items than you have available.

5. What is nC0 and nP0 ?

nC0 = 1 and nP0 = 1 for any value of n.

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How to calculate nCr and nPr?

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