01 · Explore
Fundamental Principle of Counting
The fundamental principle of counting, also known as the multiplication principle, is the bedrock of combinatorics. it allows us to determine the total number of ways multiple events can occur without listing every possibility.
If an event can occur in 'm' different ways, and following its occurrence, another event can occur in 'n' different ways, then the total number of ways the two events can occur in that specific order is m × n. This logic extends to any finite number of events. For instance, if a third event follows in 'p' ways, the total is m × n × p.
Consider a practical example: if you have 3 pairs of pants and 2 shirts, for every single pair of pants you choose, you have 2 choices of shirts. Thus, the total combinations are 3 × 2 = 6. This principle is essential when dealing with large sets where manual listing is tedious and error-prone.
Successive Choice Logic
- 1
Event 1
Choose one item from 'm' options (e.g., 3 pants).
- 2
Event 2
For each choice in Event 1, choose from 'n' options (e.g., 2 shirts).
- 3
Total Outcomes
The product of options: m × n (e.g., 3 × 2 = 6 combinations).
The multiplication principle showing how successive independent choices multiply to form the total sample space.
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Question
A person has 2 school bags, 3 tiffin boxes, and 2 water bottles. In how many ways can they carry one of each to school?
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02 · Explore
Factorial Notation
Factorial notation is a mathematical shorthand used to represent the product of the first 'n' natural numbers.
The symbol n! (read as 'n factorial') represents the product 1 × 2 × 3 × ... × n. For example, 4! = 4 × 3 × 2 × 1 = 24. By definition, 0! is always equal to 1. This notation is crucial for simplifying complex permutation and combination formulas.
A useful property of factorials is their recursive nature: n! = n × (n - 1)!. This allows us to simplify fractions easily. For example, 7! / 5! can be written as (7 × 6 × 5!) / 5!, which simplifies to 7 × 6 = 42.
Simplifying Factorial Expressions
1/8! + 1/9! = x/10!
To find x, we express all terms with the largest denominator (10!). Multiply the first term by (10 × 9) and the second by 10. This gives (90 + 10) / 10! = x / 10!. Therefore, 100 / 10! = x / 10!, so x = 100.
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Question
Evaluate 7! - 5!.
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03 · Explore
Permutations of Distinct Objects
A permutation is an arrangement of objects in a definite order. When the order of objects changes, it constitutes a new permutation.
The number of permutations of 'n' different objects taken 'r' at a time, where repetition is not allowed, is denoted by nPr. The formula is derived from the multiplication principle: the first place can be filled in 'n' ways, the second in (n-1) ways, and so on, up to the 'r-th' place which has (n - r + 1) ways.
The standard formula is nPr = n! / (n - r)!, where 0 ≤ r ≤ n. If we arrange all 'n' objects (r = n), the formula becomes nPn = n! / 0! = n!. If repetition is allowed, each of the 'r' places can be filled in 'n' ways, resulting in n^r total permutations.
| Condition | Formula | Example (n=4, r=2) |
|---|---|---|
| No Repetition | n! / (n - r)! | 4! / 2! = 12 |
| Repetition Allowed | n^r | 4² = 16 |
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How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if repetition is not allowed?
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04 · Explore
Permutations of Non-Distinct Objects
When some objects in a set are identical, the total number of unique arrangements decreases because swapping identical objects does not create a new arrangement.
If we have 'n' objects where 'p' objects are of the same kind and the rest are different, the number of unique permutations is n! / p!. This is because the 'p!' ways we could have arranged those identical objects (if they were distinct) now only count as one single arrangement.
In a more general case, if there are n objects with p1 objects of one kind, p2 of a second kind, and so on up to pk, the total permutations are given by n! / (p1! × p2! × ... × pk!). For example, in the word 'ROOT', there are 4 letters with two 'O's. The arrangements are 4! / 2! = 12.
Arranging the word ALLAHABAD
9! / (4! × 2!)
The word has 9 letters. 'A' appears 4 times and 'L' appears 2 times. Total arrangements = 362880 / (24 × 2) = 7560.
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Question
How many arrangements can be made from the letters of the word INSTITUTE?
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05 · Explore
Permutations with Constraints
Many problems require arrangements where certain items must stay together or stay apart.
When items must occur together, we treat them as a single 'block' or object. We then arrange this block along with the remaining items. Finally, we multiply by the number of ways the items inside the block can be arranged among themselves.
To find arrangements where items are 'never together', it is often easiest to calculate the total possible arrangements without restrictions and subtract the number of arrangements where they are 'always together'.
Vowels Together in DAUGHTER
6! × 3!
The word DAUGHTER has 8 letters. Vowels are A, U, E. Treat (AUE) as one object. Now we have 5 consonants + 1 block = 6 objects. These can be arranged in 6! ways. The 3 vowels inside the block can be arranged in 3! ways. Total = 720 × 6 = 4320.
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Question
In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
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06 · Explore
The Concept of Combinations
A combination is a selection of items where the order does not matter. This distinguishes it from a permutation.
In combinations, the group {A, B} is the same as {B, A}. This is common in scenarios like forming a committee, shaking hands, or drawing chords between points on a circle. Because order is ignored, the number of combinations is always less than or equal to the number of permutations for the same set.
The number of combinations of 'n' objects taken 'r' at a time is denoted by nCr. Since each combination of 'r' objects can be arranged in r! ways to create permutations, we have the relationship: nPr = nCr × r!. Rearranging this gives the formula: nCr = n! / (r! × (n - r)!).
| Feature | Permutation (nPr) | Combination (nCr) |
|---|---|---|
| Order | Important | Not Important |
| Logic | Arrangement | Selection |
| Formula | n! / (n - r)! | n! / (r! (n - r)!) |
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Question
How many chords can be drawn through 21 points on a circle?
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07 · Explore
Properties and Theorems of Combinations
Several mathematical identities simplify the calculation of combinations and reveal the symmetry of selections.
One fundamental property is symmetry: nCr = nC(n-r). This means selecting 'r' objects is mathematically equivalent to 'rejecting' (n-r) objects. For example, 10C8 is the same as 10C2, which is much easier to calculate.
Another important result is Pascal's Rule (Theorem 6): nCr + nC(r-1) = (n+1)Cr. This identity is the basis for constructing Pascal's Triangle. Additionally, if nCa = nCb, then either a = b or a + b = n.
Finding n from Equality
nC9 = nC8
Using the property that if nCa = nCb then n = a + b, we find n = 9 + 8 = 17. To find 17C17, we use the rule nCn = 1. Thus, 17C17 = 1.
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If nC8 = nC2, find nC2.
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08 · Explore
Practical Applications of Combinations
Combinations are used to solve complex selection problems involving multiple groups, such as card games or committee formations.
When selecting from multiple distinct groups, we calculate the combinations for each group separately and multiply them using the Fundamental Principle of Counting. For example, to select a team of 3 boys from 5 and 3 girls from 4, we calculate 5C3 × 4C3.
In card problems, we often deal with suits (13 cards each) or colors (26 cards each). If a condition specifies 'at least' a certain number, we must sum the combinations for all possible scenarios that meet that minimum requirement.
Committee Selection
2C1 × 3C2
To form a committee of 3 from 2 men and 3 women such that it has 1 man and 2 women: Ways to pick 1 man = 2C1 = 2. Ways to pick 2 women = 3C2 = 3. Total ways = 2 × 3 = 6.
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Question
In how many ways can a team of 11 be chosen from 17 players, if 5 are bowlers and the team must have exactly 4 bowlers?
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