Back to Course
Permutation and Probability
Fundamental Counting Principle and Factorials
The Fundamental Counting Principle (FCP) states that if one event can occur in $p$ ways and, after it has occurred, a second event can occur in $q$ ways, then the total number of ways for the sequence is the product $p \times q$.
$$n! = n \times (n-1) \times (n-2) \dots 1$$
Factorial notation represents the product of all positive integers up to $n$.
$n!$=$n$ factorial — product of all integers from $1$ to $n$(unitless)
$0!$=Zero factorial, defined as $1$ by convention(unitless)
$n = 0$
→$0! = 1$ — required for consistent formulas like $P(n, n) = \frac{n!}{0!} = n!$
Sequential Choices: FCP applies to any sequence of choices — whether the pool shrinks (without replacement) or stays the same (with replacement).
Factorial as Arrangement Count: $n!$ gives the number of ways to arrange $n$ distinct objects in a line.
Recursive Definition: $n! = n \times (n-1)!$, so $5! = 5 \times 4! = 5 \times 24 = 120$.
Simplification Shortcut: $\frac{n!}{(n-r)!}$ cancels to the product of $r$ descending factors: $\frac{8!}{5!} = 8 \times 7 \times 6 = 336$.
Permutations: When Order Matters
A Permutation is an ordered arrangement of a set of distinct objects. Changing the sequence (e.g., ABC vs BCA) produces a different permutation.
$$^nP_r = \frac{n!}{(n-r)!}$$
Calculates the number of ways to arrange $r$ objects chosen from $n$ unique objects, where order matters.
$n$=Total number of distinct items available(count)
$r$=Number of items to arrange ($r \le n$)(count)
$r = n$
→$^nP_n = n!$ — arranging all items
$r = 1$
→$^nP_1 = n$ — just choosing one item
Descending Product: $^nP_r$ equals $r$ consecutive integers multiplied: $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$.
Useful Identity: $^nP_r = n \cdot ^{n-1}P_{r-1}$ — fix one item first, then arrange the rest.
Permutations with Repeated and Circular Arrangements
When the set contains Identical Items, swapping them does not create a new arrangement. We divide $n!$ by the factorials of each group's count to remove duplicates.
$$\text{Arrangements} = \frac{n!}{n_1! \, n_2! \dots n_k!}$$
Counts distinct permutations when some objects are indistinguishable.
$n$=Total number of objects(count)
$n_i$=Number of identical objects in group $i$(count)
All objects distinct ($n_i = 1$ for all $i$)
→Reduces to $n!$ since each $n_i! = 1$
Why Divide?: If a letter appears $k$ times, those $k$ copies can be rearranged among themselves in $k!$ ways without producing a visibly new arrangement. Dividing removes this overcounting.
MISSISSIPPI Example: 11 letters with I×4, S×4, P×2, M×1 gives $\frac{11!}{4! \cdot 4! \cdot 2! \cdot 1!} = 34650$.
In a Circular Permutation, rotating all objects by the same amount does not produce a new arrangement. We fix one object's position to eliminate rotational symmetry.
$$(n-1)!$$
Number of distinct ways to arrange $n$ objects in a circle.
$(n-1)!$=Fix one object, arrange the remaining $n-1$(count)
Necklace or key ring (can flip over)
→Divide by 2: $\frac{(n-1)!}{2}$ — clockwise and anticlockwise arrangements are identical.
Why $(n-1)!$?: In a line, ABCD and BCDA are different. At a round table, they are the same rotation. Fixing one person removes this redundancy.
Necklace vs Table: A necklace can be flipped, so clockwise = anticlockwise. Divide $(n-1)!$ by 2. A table cannot be flipped, so use $(n-1)!$.
Constraint Handling: If two people must sit together at a round table, treat them as one unit. For $n$ total people: $(n-2)! \times 2!$ (fix combined unit, arrange rest, swap internal order).
Combinations: When Order Does Not Matter
A Combination is a selection of $r$ objects from $n$ distinct objects where the order of selection is irrelevant. Choosing {A, B, C} is the same as choosing {C, A, B}.
$$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$
Counts the number of ways to choose $r$ items from $n$ without regard to order.
$n$=Total number of distinct items(count)
$r$=Number of items to choose(count)
$r!$=Divides out the internal orderings of the chosen group(count)
$r = 0$ or $r = n$
→$^nC_0 = ^nC_n = 1$ — only one way to choose nothing or everything
$r = 1$
→$^nC_1 = n$ — choosing one item from $n$
Permutation vs Combination: Permutation = arrangement (order matters). Combination = selection (order doesn't matter). A committee of 3 from 10 is $^{10}C_3$, but assigning President/VP/Secretary from 10 is $^{10}P_3$.
Complementary Property: $^nC_r = ^nC_{n-r}$ — choosing $r$ items to include is equivalent to choosing $n-r$ items to exclude. Use this when $r > n/2$ to simplify: $^{12}C_{10} = ^{12}C_2 = 66$.
Pascal's Identity: $^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$ — the basis for Pascal's triangle.
Computational Shortcut: $^nC_r = \frac{n(n-1)\dots(n-r+1)}{r!}$ — multiply $r$ descending factors, then divide by $r!$. For $^{20}C_{17}$, use $^{20}C_3 = \frac{20 \times 19 \times 18}{3!} = 1140$.
Combinations are used to count selections in geometry (diagonals, triangles from vertices) and in forming groups with constraints.
Diagonals of a Polygon: Total line segments from $n$ vertices = $^nC_2$. Subtract $n$ sides to get diagonals: $^nC_2 - n = \frac{n(n-3)}{2}$.
Triangles from Vertices: Number of triangles = $^nC_3$ (choose any 3 non-collinear vertices).
Constrained Selection: To form a committee that must include 2 specific people from 8, choose the remaining members: $^{8-2}C_{5-2} = ^6C_3 = 20$.
Finding $n$ from $^nC_r$: If $^nC_a = ^nC_b$, then either $a = b$ or $a + b = n$ (by the complementary property).
Probability Fundamentals
Probability is the numerical measure of the likelihood that an event occurs, based on the ratio of favorable outcomes to total outcomes in a Sample Space of Equally Likely outcomes.
$$P(E) = \frac{n(E)}{n(S)}$$
The classical (Laplace) definition of probability for equally likely outcomes.
$n(E)$=Number of outcomes favorable to event $E$(count)
$n(S)$=Total number of outcomes in the sample space(count)
$P(E) = 0$
→Impossible event — cannot occur
$P(E) = 1$
→Certain event — guaranteed to occur
Range: $0 \le P(E) \le 1$ always. If you calculate a probability outside this range, recheck your work.
Complement Rule: $P(\bar{E}) = 1 - P(E)$ — the probability that event $E$ does NOT occur. Often easier to compute.
Sample Space Examples: Coin toss: $S = \{H, T\}$, $n(S) = 2$. Die roll: $S = \{1,2,3,4,5,6\}$, $n(S) = 6$. Two dice: $n(S) = 36$.
Two events are Mutually Exclusive if they cannot occur at the same time — their intersection is empty ($A \cap B = \emptyset$).
Disjoint Sets: If $A$ and $B$ are mutually exclusive, then $P(A \cap B) = 0$.
Example: Rolling a single die — the events 'getting 2' and 'getting 5' are mutually exclusive.
Non-Example: 'Getting an even number' and 'getting a number > 3' are NOT mutually exclusive because 4 and 6 satisfy both.
Addition and Multiplication Rules of Probability
The Addition Rule calculates the probability that at least one of two events occurs. For overlapping events, subtract the intersection to avoid double-counting.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Probability of event $A$ OR event $B$ (or both) occurring.
$P(A \cup B)$=Probability of $A$ or $B$ occurring(probability)
$P(A \cap B)$=Probability of both $A$ and $B$ occurring (overlap)(probability)
$A$ and $B$ are mutually exclusive
→$P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$
Disjoint Case: When events cannot overlap, simply add their probabilities: $P(A \cup B) = P(A) + P(B)$.
Overlapping Case: When events can co-occur, subtracting $P(A \cap B)$ prevents counting the overlap twice.
Example: Die roll — $P(\text{prime or odd}) = P(\{2,3,5\}) + P(\{1,3,5\}) - P(\{3,5\}) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{2}{3}$.
For Independent Events, the probability that both occur is the product of their individual probabilities.
$$P(A \cap B) = P(A) \times P(B)$$
Probability of event $A$ AND event $B$ both occurring, when they are independent.
$P(A)$=Probability of event $A$(probability)
$P(B)$=Probability of event $B$(probability)
Multiple independent events $A_1, A_2, \dots, A_n$
→$P(A_1 \cap A_2 \cap \dots \cap A_n) = P(A_1) \times P(A_2) \times \dots \times P(A_n)$
Independence Defined: Two events are Independent Events if the occurrence of one does not affect the probability of the other.
With vs Without Replacement: Drawing with replacement keeps events independent (pool stays the same). Drawing without replacement makes events dependent (pool changes).
Example: Tossing a coin twice — $P(\text{Head then Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Application Strategy
Worked Examples
Problem Statement
Find the number of ways to arrange the letters in the word 'BANANA'.
Permutations with identical itemsFactorials
Strategic Toolkit
$$\frac{n!}{n_1! \, n_2! \dots n_k!}$$
Step 1
Identify Counts
Total letters $n = 6$. Identify repetitions: A appears 3 times, N appears 2 times, B appears 1 time.
$n = 6, \, n_A = 3, \, n_N = 2, \, n_B = 1$
Step 2
Apply Formula
Set up the multiset permutation formula using the counts.
$$\text{Ways} = \frac{6!}{3! \cdot 2! \cdot 1!}$$
Step 3
Calculate Result
Expand factorials and simplify.
$$\frac{720}{6 \times 2 \times 1} = 60$$
Problem Statement
5 people (A, B, C, D, E) are sitting in a row. What is the probability that A and B are NOT sitting next to each other?
Complementary ProbabilityPermutation with constraints (Grouping)
Strategic Toolkit
$$P(\bar{E}) = 1 - P(E)$$
Step 1
Find Total Sample Space
Calculate total possible arrangements of 5 distinct people.
$n(S) = 5! = 120$
Step 2
Calculate 'Together' Arrangements
Treat (AB) as 1 unit. Now we have 4 units: {AB, C, D, E}. Permute these 4 units ($4!$) and account for internal swap of A and B ($2!$).
$n(E) = 4! \times 2! = 24 \times 2 = 48$
Step 3
Find Complement
Subtract 'together' count from total to find 'not together' arrangements.
$n(\bar{E}) = 120 - 48 = 72$
Step 4
Final Probability
Divide favorable outcomes by sample space.
$$P(\bar{E}) = \frac{72}{120} = \frac{3}{5}$$
Problem Statement
From 8 men and 10 women, form a committee of 7 with at least 4 women. How many ways?
CombinationsConstrained selectionCase analysis
Strategic Toolkit
$$^nC_r = \frac{n!}{r!(n-r)!}$$
Step 1
Identify Cases
At least 4 women means: Case 1 (4W, 3M), Case 2 (5W, 2M), Case 3 (6W, 1M), Case 4 (7W, 0M).
Cases: $(4,3), (5,2), (6,1), (7,0)$
Step 2
Calculate Each Case
For each case, multiply women choices by men choices.
$^{10}C_4 \cdot ^8C_3 + ^{10}C_5 \cdot ^8C_2 + ^{10}C_6 \cdot ^8C_1 + ^{10}C_7 \cdot ^8C_0$
Step 3
Compute Values
Calculate each term: $210 \times 56 + 252 \times 28 + 210 \times 8 + 120 \times 1$.
$11760 + 7056 + 1680 + 120 = 20616$
Problem Statement
A card is drawn from a standard deck of 52. What is the probability that it is a diamond or an ace?
Addition of probabilitiesOverlapping events
Strategic Toolkit
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Step 1
Identify Events
Let $A$ = diamond (13 cards), $B$ = ace (4 cards). These overlap: the ace of diamonds is in both.
$n(A) = 13, \, n(B) = 4, \, n(A \cap B) = 1$
Step 2
Apply Addition Rule
Since the events overlap, subtract the intersection.
$$P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$
Problem Statement
In how many ways can 5 men and 5 women be seated at a round table so that no two people of the same sex sit together?
Circular permutationAlternating constraint
Strategic Toolkit
$$(n-1)!$$
Step 1
Fix One Group
Fix one man to remove rotational symmetry. The remaining 4 men can be arranged in the alternate seats.
$\text{Men} = (5-1)! = 4! = 24$
Step 2
Fill Remaining Seats
The 5 gaps between men are now fixed. Place 5 women in these 5 seats.
$\text{Women} = 5! = 120$
Step 3
Total Arrangements
Multiply the two independent arrangements.
$$4! \times 5! = 24 \times 120 = 2880$$
Memorize formulas
Permutation Formula
Arranging $r$ items from $n$ distinct items in a specific order.
Formula
$$^nP_r = \frac{n!}{(n-r)!}$$
Combination Formula
Selecting $r$ items from $n$ distinct items (order irrelevant).
Formula
$$^nC_r = \frac{n!}{r!(n-r)!}$$
Repeated Items Permutation
Arrangements when some items are identical.
Formula
$$\frac{n!}{n_1! \, n_2! \dots n_k!}$$
Circular Permutation
Arranging $n$ items around a circle; $\frac{(n-1)!}{2}$ for necklaces.
Formula
$$(n-1)!$$
Classical Probability
Ratio of favorable outcomes to total equally likely outcomes.
Formula
$$P(E) = \frac{n(E)}{n(S)}$$
Complement Probability
Probability that an event does NOT occur.
Formula
$$P(\bar{E}) = 1 - P(E)$$
Addition Rule
Probability of $A$ or $B$ occurring (subtract overlap).
Formula
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Multiplication Rule (Independent)
Probability of both independent events occurring.
Formula
$$P(A \cap B) = P(A) \cdot P(B)$$
Review
1
Nine people are to be seated at a round table. If three specific people must all sit next to each other, how many unique arrangements are possible?
Ans
4320
Review
Tap to reveal
Treat the trio as one block
The three specific people must stay together, so count them as a single unit. That leaves 7 objects to arrange around the round table, giving a circular count of $(7-1)!$.
$$9-3+1=7,\quad (7-1)! = 6! = 720$$
Arrange the trio inside the block
The three people in the block can permute among themselves in $3!$ ways. Multiply that internal arrangement count by the circular arrangement count.
$$720 \times 3! = 720 \times 6 = 4320$$
2
In a race with 8 runners, what is the probability that a specific trio (Sam, Joe, and Ari) will finish in exactly that order (Sam 1st, Joe 2nd, Ari 3rd)?
Ans
$\frac{1}{336}$
Review
Tap to reveal
Count ordered top-three finishes
The order of finish matters, so use permutations rather than combinations. The total number of possible ordered top-3 outcomes from 8 runners is $^8P_3$.
$$^8P_3 = 8 \times 7 \times 6 = 336$$
Identify the single favorable ordering
Only one outcome matches Sam first, Joe second, and Ari third. The probability is therefore one favorable ordering divided by all ordered top-3 outcomes.
$$P = \frac{1}{336}$$
3
If you arrange the letters of the word 'LEVEL' at random, what is the probability that the arrangement starts and ends with 'L'?
Ans
$\frac{1}{10}$
Review
Tap to reveal
Count all distinct arrangements
The word LEVEL has repeated letters, so divide by repeated-letter factorials when counting all arrangements.
$$\text{Total} = \frac{5!}{2!\,2!} = 30$$
Fix L at both ends
The two L's must occupy the first and last positions. The middle letters are E, E, and V, which can be arranged in $\frac{3!}{2!}$ ways.
$$\text{Favorable} = \frac{3!}{2!} = 3,\quad P = \frac{3}{30} = \frac{1}{10}$$
4
If $^nC_8 = ^nC_{12}$, find $n$.
Ans
20
Review
Tap to reveal
Use the complement identity
For combinations, $^nC_r = ^nC_{n-r}$. So if $^nC_8$ equals $^nC_{12}$, the indices must be complementary values.
$$n-8 = 12$$
Solve for n
Add 8 to both sides to get the total number of items being chosen from.
$$n = 20$$
5
How many committees of 5 members can be chosen from a group of 8 persons when each committee must include 2 particular persons?
Ans
20
Review
Tap to reveal
Keep the required members fixed
Since the committee must include the two particular persons, they are already chosen. That leaves 3 more seats to fill.
$$5-2 = 3$$
Choose the remaining members
Pick the remaining 3 committee members from the other 6 people using combinations.
$$^6C_3 = \frac{6!}{3!\,3!} = 20$$
6
A necklace is to be made from 7 beads of different colours. If two particular beads (red and blue) must always be adjacent, how many distinct necklaces are possible?
Ans
120
Review
Tap to reveal
Combine the adjacent beads into one block
Red and blue must stay together, so treat them as one unit. That reduces the 7 beads to 6 necklace objects, and distinct necklaces of 6 objects are counted as $\frac{(6-1)!}{2}$.
$$\frac{(6-1)!}{2} = 60$$
Account for the internal order of the block
Inside the block, red and blue can switch places in 2 ways. Multiply the necklace count by this internal symmetry factor.
$$60 \times 2! = 120$$
7
A card is drawn from a standard deck of 52. What is the probability that it is a diamond card or an ace?
Ans
$\frac{4}{13}$
Review
Tap to reveal
Apply inclusion-exclusion
A card can be a diamond, an ace, or both, so add the two counts and subtract the overlap once.
$$P(D \cup A)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}$$
Simplify the fraction
Reduce the resulting probability to its simplest form.
$$\frac{16}{52}=\frac{4}{13}$$
8
Two dice are thrown. What is the probability that the sum of dots is 4 or 6?
Ans
$\frac{2}{9}$
Review
Tap to reveal
Count the sum-4 outcomes and sum-6 outcomes
List the ordered pairs that produce each sum. There are 3 ways to get 4 and 5 ways to get 6.
$$3 + 5 = 8$$
Divide by the total sample space
Two dice have 36 equally likely ordered outcomes, and the two sum events do not overlap.
$$P=\frac{8}{36}=\frac{2}{9}$$
9
A fair die is thrown twice. What is the probability that a prime number appears on the first throw and the number on the second throw is less than 5?
Ans
$\frac{1}{3}$
Review
Tap to reveal
Find each event probability
The first die is prime with probability $3/6$, and the second die is less than 5 with probability $4/6$.
$$P(A)=\frac{1}{2},\quad P(B)=\frac{2}{3}$$
Multiply because the throws are independent
Separate throws do not affect each other, so the joint probability is the product of the two probabilities.
$$P(A\cap B)=\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}$$
10
How many 6-digit numbers can be formed without repeating any digit from the digits 0, 1, 2, 3, 4, 5?
Ans
600
Review
Tap to reveal
Restrict the leading digit
A 6-digit number cannot begin with 0, so the first position has only 5 choices: 1 through 5.
$$5\text{ choices}$$
Arrange the remaining digits
After fixing the first digit, the other 5 digits can be arranged freely in the remaining 5 places.
$$5 \times 5! = 5 \times 120 = 600$$
11
In how many ways can a hockey team of 11 players be selected from 15 players? How many of these selections include a particular player?
Ans
$^{15}C_{11} = 1365$ total; $^{14}C_{10} = 1001$ including the player
Review
Tap to reveal
Count all 11-player teams
A team selection ignores order, so use combinations to choose 11 players from 15.
$$^{15}C_{11} = ^{15}C_4 = 1365$$
Force the particular player onto the team
If one player must be included, choose the remaining 10 players from the other 14.
$$^{14}C_{10} = ^{14}C_4 = 1001$$
12
In how many ways can 8 books including 2 on English be arranged on a shelf such that the 2 English books are never together?
Ans
30240
Review
Tap to reveal
Count the complement
It is easier to subtract the arrangements where the two English books are together from all possible shelf arrangements.
$$8! - 7!\times 2!$$
Evaluate the subtraction
Compute the total arrangements and the together cases, then subtract.
$$40320 - 10080 = 30240$$
Question 1 of 12
0:00
1
Nine people are to be seated at a round table. If three specific people must all sit next to each other, how many unique arrangements are possible?