Welcome, future engineers and mathematicians! Today, we're diving deep into two fundamental concepts in Permutations and Combinations:
P(n,r) and
C(n,r). These aren't just formulas; they represent powerful ways to count possibilities, which is the cornerstone of probability and advanced mathematics. We'll start from the absolute basics, understand the 'why' behind the formulas, and then see how they are applied in various scenarios, including those relevant for
JEE Advanced.
Before we jump into P(n,r) and C(n,r), let's quickly recall the
Fundamental Principle of Counting, which forms the bedrock of these concepts.
1. Recap: The Fundamental Principle of Counting (FPC)
The FPC states that if an event can occur in 'm' ways and another independent event can occur in 'n' ways, then:
- Multiplication Principle: Both events can occur in sequence in m * n ways. (Think of choosing a shirt AND a pant).
- Addition Principle: Either one of the events can occur in m + n ways. (Think of choosing a shirt OR a pant, but not both at the same time).
P(n,r) and C(n,r) are essentially sophisticated applications of the Multiplication Principle.
2. Permutations: P(n,r) - When Order Matters!
Imagine you have a set of distinct objects, and you want to arrange some of them. The act of arranging implies that the
order in which you place them is important. This is where permutations come in.
2.1. What is a Permutation?
A
permutation is an arrangement of a set of objects in a definite order.
Think of it like arranging books on a shelf, seating people in chairs, or forming a sequence of digits. In all these cases, changing the order creates a different outcome.
2.2. The Meaning of P(n,r)
P(n,r), often written as $^n P_r$, represents the number of ways to arrange (or permute) 'r' distinct objects chosen from a set of 'n' distinct objects.
*
'n' is the total number of distinct objects available.
*
'r' is the number of objects you are choosing and arranging at a time.
* The condition is always $0 le r le n$.
2.3. Derivation of the Formula for P(n,r)
Let's understand how the formula comes about using the Multiplication Principle.
Suppose we have 'n' distinct objects, and we want to arrange 'r' of them in 'r' positions.
1.
For the 1st position: We have 'n' choices (any of the 'n' objects).
2.
For the 2nd position: Since one object is already placed, we are left with (n-1) choices.
3.
For the 3rd position: We have (n-2) choices remaining.
4. ...
5. This pattern continues until we fill the 'r'-th position.
For the r-th position: We will have used (r-1) objects. So, the number of choices remaining will be $n - (r-1) = n - r + 1$.
By the Multiplication Principle, the total number of ways to arrange 'r' objects from 'n' is the product of the number of choices for each position:
$$ P(n,r) = n imes (n-1) imes (n-2) imes dots imes (n-r+1) $$
To express this using factorials (which are super useful in combinatorics), we can multiply and divide by $(n-r)!$:
$$ P(n,r) = frac{n imes (n-1) imes (n-2) imes dots imes (n-r+1) imes (n-r)!}{(n-r)!} $$
The numerator is simply $n!$.
Thus, the formula for P(n,r) is:
$$ oxed{P(n,r) = frac{n!}{(n-r)!}} $$
2.4. Special Case: Permutations of all n objects
If we want to arrange all 'n' distinct objects, then $r = n$.
$$ P(n,n) = frac{n!}{(n-n)!} = frac{n!}{0!} $$
Since $0! = 1$ (by definition, to make mathematical formulas consistent),
$$ P(n,n) = n! $$
This means there are $n!$ ways to arrange 'n' distinct objects.
2.5. Examples of Permutations
Example 1 (CBSE Level):
In how many ways can 5 distinct books be arranged on a shelf?
Solution: Here, $n=5$ (total distinct books) and $r=5$ (we are arranging all of them).
So, $P(5,5) = 5! = 5 imes 4 imes 3 imes 2 imes 1 = 120$ ways.
Example 2 (JEE Mains Level):
How many different 3-letter words (meaningful or meaningless) can be formed using the letters of the word 'MATHS', if no letter is repeated?
Solution:
The letters in 'MATHS' are M, A, T, H, S. There are 5 distinct letters ($n=5$).
We need to form 3-letter words, which means we are choosing and arranging 3 letters at a time ($r=3$).
Since the order of letters matters (e.g., MAT is different from TAM), this is a permutation problem.
$$ P(5,3) = frac{5!}{(5-3)!} = frac{5!}{2!} = frac{5 imes 4 imes 3 imes 2 imes 1}{2 imes 1} = 5 imes 4 imes 3 = 60 $$
So, 60 different 3-letter words can be formed.
Example 3 (Advanced JEE Type):
In how many ways can 5 boys and 4 girls be arranged in a row so that no two girls sit together?
Solution:
This is a classic 'gap method' problem, which uses permutations.
First, arrange the boys. Since there are 5 boys, they can be arranged in $P(5,5) = 5! = 120$ ways.
_B_B_B_B_B_
When the 5 boys are arranged, they create 6 possible 'gaps' (marked by underscores) where the girls can be seated so that no two girls are together.
Now, we need to choose 4 of these 6 gaps for the 4 girls and arrange the girls in those chosen gaps.
This is $P(6,4)$ because order of girls matters (Girl1 in gap1, Girl2 in gap2 is different from Girl2 in gap1, Girl1 in gap2).
$$ P(6,4) = frac{6!}{(6-4)!} = frac{6!}{2!} = frac{6 imes 5 imes 4 imes 3 imes 2 imes 1}{2 imes 1} = 6 imes 5 imes 4 imes 3 = 360 $$
By the Multiplication Principle, the total number of arrangements is $5! imes P(6,4) = 120 imes 360 = 43200$.
3. Combinations: C(n,r) - When Order Does NOT Matter!
Now, let's consider situations where the order of selection doesn't change the outcome. If you pick a group of people for a committee, it doesn't matter who was picked first or last; the committee remains the same. This is the realm of combinations.
3.1. What is a Combination?
A
combination is a selection of objects where the order of selection does not matter.
Think of it like choosing a team from a group of players, selecting a few items from a menu, or picking cards from a deck. The set of chosen items is what matters, not the sequence in which they were chosen.
3.2. The Meaning of C(n,r)
C(n,r), often written as $^n C_r$ or $inom{n}{r}$, represents the number of ways to choose (or select) 'r' distinct objects from a set of 'n' distinct objects, without regard to their order.
*
'n' is the total number of distinct objects available.
*
'r' is the number of objects you are choosing at a time.
* The condition is always $0 le r le n$.
3.3. Derivation of the Formula for C(n,r)
The derivation of combinations comes directly from permutations.
Let's consider selecting 'r' objects from 'n' objects.
1. Suppose we select 'r' objects from 'n' objects. The number of ways to do this is $C(n,r)$.
2. Once we have selected these 'r' objects, we can arrange them among themselves in $r!$ ways.
So, if we first choose 'r' objects (in $C(n,r)$ ways) and then arrange them (in $r!$ ways), what do we get? We get the total number of arrangements of 'r' objects from 'n' objects, which is $P(n,r)$.
Therefore, we can establish the relationship:
$$ P(n,r) = C(n,r) imes r! $$
From this, we can solve for $C(n,r)$:
$$ C(n,r) = frac{P(n,r)}{r!} $$
Now, substitute the formula for $P(n,r) = frac{n!}{(n-r)!}$:
$$ C(n,r) = frac{frac{n!}{(n-r)!}}{r!} $$
Thus, the formula for C(n,r) is:
$$ oxed{C(n,r) = frac{n!}{r!(n-r)!}} $$
3.4. Properties of Combinations
A very useful property is:
$$ C(n,r) = C(n, n-r) $$
This means choosing 'r' items from 'n' is the same as choosing 'n-r' items to leave behind.
Example: Choosing 3 students for a committee from 10 is the same as choosing 7 students NOT for the committee from 10. $C(10,3) = C(10,7)$. This property can often simplify calculations.
3.5. Examples of Combinations
Example 1 (CBSE Level):
In how many ways can a committee of 3 students be selected from a group of 10 students?
Solution:
Here, $n=10$ (total students) and $r=3$ (students to be selected).
Since the order of selecting students for a committee does not matter (i.e., selecting A then B then C is the same as selecting B then A then C), this is a combination problem.
$$ C(10,3) = frac{10!}{3!(10-3)!} = frac{10!}{3!7!} = frac{10 imes 9 imes 8 imes 7!}{3 imes 2 imes 1 imes 7!} = frac{10 imes 9 imes 8}{3 imes 2 imes 1} = 10 imes 3 imes 4 = 120 $$
So, 120 different committees can be formed.
Example 2 (JEE Mains Level):
A box contains 5 different red balls and 3 different blue balls. In how many ways can 3 red balls and 2 blue balls be selected?
Solution:
This problem involves two independent selections, so we'll use the Multiplication Principle.
1.
Select 3 red balls from 5 red balls: Since the order of selection doesn't matter (you just want the group of 3 balls), this is a combination.
$C(5,3) = frac{5!}{3!(5-3)!} = frac{5!}{3!2!} = frac{5 imes 4}{2 imes 1} = 10$ ways.
2.
Select 2 blue balls from 3 blue balls: Similarly, this is a combination.
$C(3,2) = frac{3!}{2!(3-2)!} = frac{3!}{2!1!} = frac{3 imes 2}{2 imes 1} = 3$ ways.
By the Multiplication Principle, the total number of ways to select 3 red balls AND 2 blue balls is $10 imes 3 = 30$ ways.
Example 3 (Advanced JEE Type):
From a group of 7 men and 6 women, 5 persons are to be selected to form a committee so that at least 3 men are in the committee. In how many ways can this be done?
Solution:
"At least 3 men" means the committee can have:
- 3 men and 2 women
- 4 men and 1 woman
- 5 men and 0 women
We will calculate the number of ways for each case and then add them (Addition Principle).
1.
Case 1: 3 Men and 2 Women
Number of ways to choose 3 men from 7: $C(7,3) = frac{7!}{3!4!} = frac{7 imes 6 imes 5}{3 imes 2 imes 1} = 35$
Number of ways to choose 2 women from 6: $C(6,2) = frac{6!}{2!4!} = frac{6 imes 5}{2 imes 1} = 15$
Total ways for Case 1 = $35 imes 15 = 525$
2.
Case 2: 4 Men and 1 Woman
Number of ways to choose 4 men from 7: $C(7,4) = C(7,3) = 35$ (using $C(n,r)=C(n,n-r)$)
Number of ways to choose 1 woman from 6: $C(6,1) = 6$
Total ways for Case 2 = $35 imes 6 = 210$
3.
Case 3: 5 Men and 0 Women
Number of ways to choose 5 men from 7: $C(7,5) = C(7,2) = frac{7 imes 6}{2 imes 1} = 21$
Number of ways to choose 0 women from 6: $C(6,0) = 1$ (There's only one way to choose nothing)
Total ways for Case 3 = $21 imes 1 = 21$
Total number of ways = Ways(Case 1) + Ways(Case 2) + Ways(Case 3)
Total ways = $525 + 210 + 21 = 756$.
4. Permutations vs. Combinations: The Crucial Distinction
This is the most critical part of solving problems involving P(n,r) and C(n,r). Always ask yourself:
"Does the order of selection or arrangement matter?"
Feature |
Permutations (P(n,r)) |
Combinations (C(n,r)) |
|---|
Meaning |
Arrangement of 'r' objects from 'n' where order is important. |
Selection of 'r' objects from 'n' where order is NOT important. |
Keyword Clues |
Arrange, order, sequence, position, distinct roles, form words/numbers, first/second/third place. |
Select, choose, pick, group, committee, team, set, sample, collection. |
Formula |
$P(n,r) = frac{n!}{(n-r)!}$ |
$C(n,r) = frac{n!}{r!(n-r)!}$ |
Relationship |
$P(n,r) = C(n,r) imes r!$ |
$C(n,r) = frac{P(n,r)}{r!}$ |
Analogy |
A locker combination (e.g., 4-5-6 is different from 6-5-4). |
Choosing toppings for a pizza (pepperoni, mushroom, onion is same as mushroom, onion, pepperoni). |
Result |
Generally a larger number (more ways to arrange than to just pick). |
Generally a smaller number (grouping reduces possibilities). |
4.1. JEE Focus: Problem-Solving Strategy
For JEE, you'll encounter problems that require careful analysis to determine if it's a permutation, combination, or a mix of both.
1.
Read Carefully: Identify what needs to be counted โ arrangements or selections?
2.
Distinct vs. Identical Objects: P(n,r) and C(n,r) as defined here are for distinct objects. If objects are identical, the formulas change (e.g., permutations with repetitions), which is another topic.
3.
Order Matters? This is the key question.
* If yes, it's a Permutation.
* If no, it's a Combination.
4.
Break Down Complex Problems: Many JEE problems involve multiple steps. Use the Multiplication Principle for sequential events (e.g., "select this AND then select that") and the Addition Principle for mutually exclusive alternatives (e.g., "this OR that case").
5.
Consider Constraints: "At least," "at most," "no two together," "always together," etc., will guide how you apply P(n,r) and C(n,r).
By mastering the true meaning behind P(n,r) and C(n,r), you're not just memorizing formulas; you're developing a powerful analytical tool to tackle a wide range of counting problems, which is essential for success in competitive exams like the JEE. Keep practicing, and always question whether order is the silent determinant in your problem!