Quiz 3

Week 6: Permutations & Combinations

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2026-09-12T00:00:00.000Z
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Permutations nPr, combinations nCr, permutations with repeated objects, circular arrangements, and selection strategies. # Statistics for Data Science I — Week 6: Permutations & Combinations IITM’s Week 6 continues counting with **permutations and combinations**.

Statistics for Data Science I — Week 6: Permutations & Combinations

IITM’s Week 6 continues counting with permutations and combinations.

TL;DR

The central question is:
Does order matter?
If yes:
Permutation\boxed{\text{Permutation}}
If no:
Combination\boxed{\text{Combination}}
The core formulas are:
nPr=n!(n−r)!\boxed{ {}^nP_r=\frac{n!}{(n-r)!} }
and
nCr=n!r!(n−r)!\boxed{ {}^nC_r=\frac{n!}{r!(n-r)!} }

1. Permutation

A permutation is an ordered arrangement.
Example: choose 2 people from A, B, C for:
  • president
  • vice-president
Then:
(A,B)≠(B,A)(A,B)\neq(B,A)
because the roles differ.
So order matters.

2. Permutation formula

Selecting and arranging rr objects from nn distinct objects:
nPr=n!(n−r)!\boxed{ {}^nP_r= \frac{n!}{(n-r)!} }

3. Why the formula works

First position:
nn
choices.
Second:
n−1.n-1.
Continue for rr positions:
n(n−1)(n−2)⋯(n−r+1).n(n-1)(n-2)\cdots(n-r+1).
That equals:
n!(n−r)!.\frac{n!}{(n-r)!}.

4. Example

Arrange 3 people from 8 people in three distinct roles.
8P3=8!5!{}^8P_3 = \frac{8!}{5!} =8⋅7⋅6=8\cdot7\cdot6 336\boxed{336}

5. Permuting all objects

If:
r=n,r=n,
then:
nPn=n!0!.{}^nP_n = \frac{n!}{0!}.
Since:
0!=1,0!=1,
we get:
nPn=n!\boxed{ {}^nP_n=n! }

6. Combination

A combination is a selection where order does not matter.
Suppose we choose 2 people from:
A,B,C.A,B,C.
For a committee:
{A,B}={B,A}.\{A,B\}=\{B,A\}.
So these should count as one outcome.

7. Combination formula

Choosing rr objects from nn distinct objects:
nCr=(nr)=n!r!(n−r)!\boxed{ {}^nC_r= \binom nr = \frac{n!}{r!(n-r)!} }

8. Why divide by r!r!?

Start with permutations:
nPr.{}^nP_r.
Each unordered group of rr selected objects has been counted:
r!r!
times because those objects can be arranged internally in r!r! orders.
Therefore:
nCr=nPrr!.{}^nC_r = \frac{{}^nP_r}{r!}.
So:
nPr=nCr r!\boxed{ {}^nP_r = {}^nC_r\,r! }

9. Example

Choose 3 students from 8 for a committee.
8C3=8!3!5!.{}^8C_3 = \frac{8!}{3!5!}.
Cancel:
=8⋅7⋅63⋅2⋅1= \frac{8\cdot7\cdot6}{3\cdot2\cdot1} =56.=56.
Thus:
56\boxed{56}

10. Same people, different problem

This distinction is critical.

Committee of 3 from 8

Order doesn't matter:
8C3=56.{}^8C_3=56.

President, VP, secretary from 8

Order/role matters:
8P3=336.{}^8P_3=336.
Same people. Different structure.

11. Fast decision test

Ask:
If I swap two selected objects, does the outcome change?
If yes:
Permutation\boxed{\text{Permutation}}
If no:
Combination\boxed{\text{Combination}}

12. Combination symmetry

A useful identity:
(nr)=(nn−r)\boxed{ \binom nr=\binom n{n-r} }
Why?
Choosing rr objects to include is equivalent to choosing:
n−rn-r
objects to exclude.
Example:
(103)=(107).\binom{10}{3} = \binom{10}{7}.

13. Extreme combination values

(n0)=1\boxed{ \binom n0=1 }
There is exactly one way to choose nothing.
Also:
(nn)=1\boxed{ \binom nn=1 }
There is exactly one way to choose everything.
And:
(n1)=n\boxed{ \binom n1=n }

14. Example: choose 2

(n2)=n!2!(n−2)!\binom n2 = \frac{n!}{2!(n-2)!}
so:
(n2)=n(n−1)2\boxed{ \binom n2= \frac{n(n-1)}2 }
This is the same formula that appeared for the number of edges in a complete undirected graph.
Nice connection to Maths 1.

15. Pascal identity

An important identity:
(nr)=(n−1r)+(n−1r−1)\boxed{ \binom nr = \binom{n-1}{r} + \binom{n-1}{r-1} }
Intuition: distinguish one special object.
Either:
  • it is not chosen
  • it is chosen
Those cases are disjoint and exhaustive.

16. Example of Pascal identity

(53)=(43)+(42).\binom53 = \binom43+\binom42.
Numerically:
10=4+6.10=4+6.
Correct.

17. Pascal triangle

Combination numbers form Pascal's triangle:
text
        1
      1   1
    1   2   1
  1   3   3   1
1   4   6   4   1
Each interior number equals the sum of the two directly above it.

18. Sum of one row

For row corresponding to nn:
∑r=0n(nr)=2n\boxed{ \sum_{r=0}^{n} \binom nr = 2^n }
Why?
Every element of an nn-element set is either:
  • selected
  • not selected
So total subsets:
2n.2^n.
This reconnects to the power-set result from Maths 1.

19. Binomial theorem connection

The coefficients in:
(a+b)n(a+b)^n
are combinations:
(a+b)n=∑r=0n(nr)an−rbr\boxed{ (a+b)^n = \sum_{r=0}^{n} \binom nr a^{n-r}b^r }
Example:
(a+b)3=a3+3a2b+3ab2+b3.(a+b)^3 = a^3+3a^2b+3ab^2+b^3.
Coefficients:
1,3,3,1.1,3,3,1.

20. Repetition allowed: ordered selections

Suppose you choose rr items, each time from nn possibilities, and repetition is allowed.
If order matters:
nr\boxed{ n^r }
Example: 4-digit PIN:
104.10^4.

21. Repetition forbidden: ordered selections

If repetition is not allowed and order matters:
nPr\boxed{ {}^nP_r }
Example: 4 distinct digits chosen from 10:
10P4.{}^{10}P_4.

22. Repetition forbidden: unordered selections

If repetition is forbidden and order does not matter:
nCr\boxed{ {}^nC_r }
Example: choose 4 students from 10:
(104).\binom{10}{4}.

23. Core counting grid

Order matters?Repetition?Count
yesyesnrn^r
yesnonPr{}^nP_r
nononCr{}^nC_r
The "no order + repetition allowed" case exists too, but it uses a different formula and may be beyond the immediate core depending on course coverage.

24. Combinations with repetition

If you choose rr items from nn categories, order does not matter, and repetition is allowed, the count is:
(n+r−1r)\boxed{ \binom{n+r-1}{r} }
This is often called stars and bars.
Treat it as supplementary unless emphasized in your lectures.

25. Stars and bars intuition

Suppose we need nonnegative integer solutions to:
x1+x2+x3=5.x_1+x_2+x_3=5.
Represent 5 identical objects as:
text
*****
Use 2 separators:
text
**|*|**
meaning:
x1=2,x2=1,x3=2.x_1=2,\quad x_2=1,\quad x_3=2.
There are:
5+3−1=75+3-1=7
positions containing:
  • 5 stars
  • 2 bars
So number of solutions:
(72).\binom72.

26. Arrangements with repeated identical objects

Suppose there are:
nn
objects with repetitions:
n1,n2,…,nkn_1,n_2,\dots,n_k
where:
n1+⋯+nk=n.n_1+\cdots+n_k=n.
Distinct permutations:
n!n1!n2!⋯nk!\boxed{ \frac{n!}{n_1!n_2!\cdots n_k!} }

27. Example: BANANA

BANANA has:
66
letters.
Counts:
  • A: 3
  • N: 2
  • B: 1
Distinct arrangements:
6!3!2!\frac{6!}{3!2!} =72012= \frac{720}{12} 60\boxed{60}

28. Why divide by repeated factorials

If identical copies were artificially labeled, swapping them would produce fake "new" permutations.
We divide by all such internal rearrangements:
n1!n2!⋯nk!.n_1!n_2!\cdots n_k!.

29. Circular permutations

For nn distinct objects arranged around a circle, rotationally equivalent arrangements are usually treated as the same.
Then:
(n−1)!\boxed{ (n-1)! }

30. Why circular arrangements use (n−1)!(n-1)!

For linear arrangements:
n!.n!.
But each circular arrangement can be rotated into:
nn
linear descriptions.
Therefore:
n!n=(n−1)!.\frac{n!}{n} = (n-1)!.

31. Example

6 people around a round table:
(6−1)!=5!=120.(6-1)! = 5! = 120.

32. Selection with mandatory members

Suppose a committee of 4 is chosen from 10 people, and Alice must be included.
Alice is fixed.
Choose the remaining:
33
members from:
9.9.
So:
(93)\boxed{ \binom93 }

33. Selection excluding someone

Committee of 4 from 10, but Alice cannot be selected.
Then effectively choose:
44
from remaining:
9.9.
So:
(94)\boxed{ \binom94 }

34. At least one special member

Suppose 4 people are chosen from 10, including 3 women and 7 men.
How many committees contain at least one woman?
Direct counting has several cases:
  • 1 woman
  • 2 women
  • 3 women
Complement is easier.
Total committees:
(104).\binom{10}{4}.
No women:
(74).\binom74.
So:
(104)−(74)\boxed{ \binom{10}{4}-\binom74 }

35. Exactly kk special members

Suppose choose 5 people from:
  • 6 women
  • 8 men
with exactly 2 women.
Choose women:
(62).\binom62.
Choose men:
(83).\binom83.
Multiply:
(62)(83)\boxed{ \binom62\binom83 }
Because both choices must happen.

36. Why multiply combinations here?

The choices are sequential independent stages:
  1. choose women
  2. choose men
Each resulting pair of selections creates one committee.
So use the multiplication principle.

37. At most / at least language

These phrases matter.

Exactly 2

Only:
2.2.

At least 2

2,3,4,…2,3,4,\dots

At most 2

0,1,2.0,1,2.
A major source of mistakes is reading these casually.

38. Example: at most 2 women

Committee of 4 from 5 women and 6 men.
At most 2 women means:
  • 0 women
  • 1 woman
  • 2 women
Count:
(50)(64)+(51)(63)+(52)(62).\binom50\binom64 + \binom51\binom63 + \binom52\binom62.

39. Block method for arrangements

Suppose 6 people line up and A/B must stand together.
Treat A/B as one block.
Then we have:
  • block AB
  • 4 other people
So:
5!5!
block arrangements.
A and B can internally swap:
2!2!
Thus:
5!⋅2\boxed{ 5!\cdot2 }

40. Multiple blocks

Suppose:
  • A and B must be together
  • C and D must be together
among 6 people.
Treat:
[AB],[CD],E,F[AB],[CD],E,F
as 4 objects.
Arrange blocks:
4!.4!.
Internal orders:
2!⋅2!.2!\cdot2!.
Total:
4!⋅2⋅2\boxed{ 4!\cdot2\cdot2 }

41. "Not together"

Often:
not together=all−together\boxed{ \text{not together} = \text{all} - \text{together} }
For 6 people, A and B not together:
Total:
6!.6!.
Together:
5!⋅2.5!\cdot2.
Thus:
6!−2(5!)\boxed{ 6!-2(5!) }

42. Fixed positions

Suppose 7 people line up and A must stand at one end.
A has:
22
possible ends.
Remaining 6 people:
6!6!
arrangements.
Total:
2⋅6!\boxed{ 2\cdot6! }

43. A and B at the ends

If A and B must occupy the two ends:
Arrange A and B at ends:
2!2!
ways.
Arrange remaining:
(n−2)!(n-2)!
ways.
Total:
2(n−2)!\boxed{ 2(n-2)! }

44. No two special objects adjacent

A useful gap method.
Suppose:
  • 4 men
  • 3 women
  • no two women adjacent
First arrange men:
4!.4!.
This creates gaps:
text
_ M _ M _ M _ M _
Number of gaps:
5.5.
Choose 3 gaps for women:
(53).\binom53.
Arrange 3 women:
3!.3!.
Total:
4!(53)3!\boxed{ 4!\binom53 3! }

45. Why gap method works

Placing at most one special object in each gap guarantees no two are adjacent.
This is a general restriction technique worth remembering.

46. Probability bridge

Suppose 5 cards are chosen from 52, with order irrelevant.
Total possible hands:
(525)\boxed{ \binom{52}{5} }
This is why combinations are central to probability.
If every hand is equally likely:
P(E)=favorable combinations(525).P(E) = \frac{\text{favorable combinations}} {\binom{52}{5}}.

47. Example: exactly 2 special objects

Suppose a set has:
  • 4 special items
  • 6 ordinary items
Choose 3.
How many selections contain exactly 2 special items?
Choose:
22
from the 4 special:
(42).\binom42.
Choose:
11
from the 6 ordinary:
(61).\binom61.
Thus:
(42)(61)\boxed{ \binom42\binom61 }

48. Complement with combinations

Suppose we choose 3 people from 10 and want at least one of A or B.
Total:
(103).\binom{10}{3}.
Committees containing neither A nor B:
(83).\binom83.
So desired:
(103)−(83)\boxed{ \binom{10}{3}-\binom83 }

49. Exactly one of A and B

Two disjoint cases:
  • A included, B excluded
  • B included, A excluded
For committee size 3 from 10:
Choose which one of A/B:
22
ways.
Then choose remaining 2 members from the other 8 people:
(82).\binom82.
So:
2(82)\boxed{ 2\binom82 }

50. Both A and B included

For committee size 4 from 10:
Fix A and B.
Choose remaining:
22
from other:
8.8.
Count:
(82)\boxed{ \binom82 }

51. Neither A nor B

Choose all 4 from remaining 8:
(84)\boxed{ \binom84 }
These "both / exactly one / neither" patterns are common.

52. Permutation with restrictions example

How many 5-digit numbers can be formed using digits:
1,2,3,4,51,2,3,4,5
exactly once?
All digits are distinct and order matters:
5!=120.5!=120.

53. Same problem with first digit restriction

How many start with an even digit?
Even first digit:
2 or 42\text{ or }4
so:
22
choices.
Remaining:
4!4!
orders.
Total:
2⋅4!=48.2\cdot4!=48.

54. Same problem ending in odd digit

Odd choices:
1,3,51,3,5
so:
33
choices for final digit.
Remaining:
4!4!
arrangements.
Thus:
72\boxed{72}
Common mistake: using permutation for a committee
Committee members have no roles unless stated.
So:
(nr)\binom nr
not:
nPr.{}^nP_r.
Common mistake: using combination for ranks
Gold, silver, bronze are ordered positions.
So:
nP3{}^nP_3
not:
(n3).\binom n3.
Common mistake: dividing by
r!r! twice
Combination already includes the:
r!r!
division.
Do not compute:
(nr)/r!\binom nr/r!
again unless a separate symmetry genuinely requires it.
Common mistake: forgetting internal block arrangements
If A and B must be together, treating them as one block gives positions of the block but not internal order.
Usually multiply by:
2!2!
for:
ABAB
or:
BA.BA.
Common mistake: cases overlap
If solving by cases, cases must ideally be mutually exclusive.
For example:
  • exactly 1 woman
  • exactly 2 women
do not overlap.
So counts can be added safely.
But:
  • at least 1 woman
  • at least 2 women
overlap heavily.
Adding them directly double-counts.
Common mistake: forgetting impossible cases
Suppose choose 5 people with exactly 4 women when only 3 women exist.
That case contributes:
0.0.
Always check feasibility before applying formulas.

61. Simplifying combinations efficiently

Example:
(1002).\binom{100}{2}.
Do not expand 100!100!.
Use:
(n2)=n(n−1)2.\binom n2 = \frac{n(n-1)}2.
So:
(1002)=100⋅992=4950.\binom{100}{2} = \frac{100\cdot99}{2} = 4950.

62. Another simplification

(2018)\binom{20}{18}
Use symmetry:
(2018)=(202).\binom{20}{18} = \binom{20}{2}.
Thus:
=190.= 190.
Much easier.

63. Master decision tree

text
What am I counting?
│
├── Sequence / roles / ranks?
│      │
│      ├── repetition allowed → n^r
│      └── repetition forbidden → nPr
│
└── Group / committee / subset?
       │
       ├── repetition forbidden → nCr
       └── repetition allowed → stars-and-bars style
Then separately ask:
  • restrictions?
  • must include/exclude?
  • together/apart?
  • exactly/at least/at most?

64. Core formula sheet

Factorial

n!=n(n−1)⋯1\boxed{ n!=n(n-1)\cdots1 }

Permutation

nPr=n!(n−r)!\boxed{ {}^nP_r = \frac{n!}{(n-r)!} }

Combination

nCr=(nr)=n!r!(n−r)!\boxed{ {}^nC_r = \binom nr = \frac{n!}{r!(n-r)!} }

Relation

nPr=nCr r!\boxed{ {}^nP_r = {}^nC_r\,r! }

Symmetry

(nr)=(nn−r)\boxed{ \binom nr=\binom n{n-r} }

Repeated arrangements

n!n1!n2!⋯nk!\boxed{ \frac{n!}{n_1!n_2!\cdots n_k!} }

Circular permutation

(n−1)!\boxed{ (n-1)! }

65. Week 5 → Week 6 connection

Week 5 gave the fundamental counting principles:
addandmultiply.\text{add} \quad\text{and}\quad \text{multiply}.
Week 6 packages recurring counting patterns into reusable formulas:
ordered selection→nPr\boxed{ \text{ordered selection} \rightarrow {}^nP_r } unordered selection→nCr\boxed{ \text{unordered selection} \rightarrow {}^nC_r }
These become essential once probability begins.

66. What you should actually master

By the end of Stats 1 Week 6, you should be able to:
  1. Decide whether order matters.
  2. Compute permutations.
  3. Compute combinations.
  4. Explain why combinations divide by r!r!.
  5. Use: nPr=nCr r!.{}^nP_r={}^nC_r\,r!.
  6. Use symmetry: (nr)=(nn−r).\binom nr=\binom n{n-r}.
  7. Solve committee and ranking problems.
  8. Handle required/excluded members.
  9. Handle exactly / at least / at most restrictions.
  10. Use complement counting.
  11. Count arrangements with repeated identical objects.
  12. Solve basic together/not-together arrangements.
  13. Use block and gap methods when appropriate.
  14. Avoid overcounting.
If those are comfortable, Stats 1 Week 6 is under control.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Week 6 Diagnostic: In how many ways can a committee of 3 members be selected from a group of 7 candidates? :: Answer: 35 — C(7, 3) = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35.
  • Week 6 Diagnostic: In how many ways can 5 distinct books be arranged on a single shelf? :: Answer: 120 — P(5, 5) = 5! = 120.

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

In how many ways can a committee of 3 members be selected from a group of 7 candidates?
NAT
3 Unit Assessment
2

Question 2

In how many ways can 5 distinct books be arranged on a single shelf?
NAT
3 Unit Assessment

Active Recall Session

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