IIT JEE Main Maths -Unit 4- Permutations (nPr)- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 4- Permutations (nPr) – Study Notes – New syllabus
IIT JEE Main Maths -Unit 4- Permutations (nPr) – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Permutations — Definition, Formulae, and Special Cases
- Combinations — Definition, Formulae, and Properties (Including Selection with Restrictions)
Permutations — Definition, Formulae, and Special Cases
A permutation is an arrangement of some or all of a given number of distinct objects, taking into account the order of arrangement.
In simple words, permutation means the number of possible arrangements when order matters.
Formula for Number of Permutations
If \( n \) distinct objects are arranged taking \( r \) at a time, the number of permutations is given by:
$ P(n, r) = nP_r = \dfrac{n!}{(n – r)!}$
Where:
- \( n! = n \times (n – 1) \times (n – 2) \times \dots \times 1 \)
- \( n! = 1 \) if \( n = 0 \)
Important Cases
- When all objects are taken: \( P(n, n) = n! \)
- When no object is taken: \( P(n, 0) = 1 \)
Permutations of Distinguishable Objects
If there are \( n \) distinct objects in which:
- \( p_1 \) are alike of one kind,
- \( p_2 \) are alike of another kind, etc.
Then, the total number of permutations is:
$ \text{Total permutations} = \dfrac{n!}{p_1! \, p_2! \, p_3! \dots} $
Permutations with Repetition
If each of \( n \) distinct objects can be repeated \( r \) times, then the total number of permutations is:
$n^r$
Circular Permutations
When objects are arranged around a circle, rotations of the same arrangement are considered identical.
| Type | Formula | Remarks |
|---|---|---|
| Circular permutation of \( n \) distinct objects | \( (n – 1)! \) | Rotation considered same |
| Circular permutation (clockwise & anticlockwise same) | \( \dfrac{(n – 1)!}{2} \) | Used in necklace-type problems |
Special Results
- Number of ways to arrange \( n \) distinct objects = \( n! \)
- Arranging \( n \) objects in a line when some are identical = \( \dfrac{n!}{p_1!p_2! \dots} \)
- If the first or last position is fixed, reduce \( n \) by 1.
Example
How many 3-letter words can be formed using the letters of the word “DELTA” without repetition?
▶️ Answer / Explanation
Total letters = 5, we need to choose 3 and arrange them.
\( 5P_3 = \dfrac{5!}{(5 – 3)!} = \dfrac{120}{2} = 60 \)
Answer: 60 possible 3-letter words.
Example
How many different words can be formed with the letters of the word “BANANA”?
▶️ Answer / Explanation
Total letters = 6, with 3 A’s and 2 N’s identical.
Number of distinct arrangements:
\( \dfrac{6!}{3! \times 2!} = \dfrac{720}{12} = 60 \)
Answer: 60 distinct arrangements.
Example
In how many ways can 6 people sit around a circular table if two particular persons must always sit together?
▶️ Answer / Explanation
Step 1: Treat the two persons who must sit together as one block.
This reduces total persons to \( (6 – 1) = 5 \) effective people.
Step 2: Circular permutations = \( (5 – 1)! = 4! = 24 \)
Step 3: The two persons in the block can exchange seats among themselves = \( 2! = 2 \) ways.
Total arrangements: \( 24 \times 2 = 48 \)
Answer: 48 possible seating arrangements.
Notes and Study Materials
- Concepts of Permutations & Combinations
- Permutations & Combinations Master File
- Permutations & Combinations Revision Notes
- Permutations & Combinations Formulae
- Permutations & Combinations Reference Book
- Permutations & Combinations Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,” Permutations & Combinations” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
The fundamental principle of counting. Permutation as an arrangement and combination as selection . The meaning of P (n,r) and C (n,r). Simple applications.
IITian Academy Notes for IIT JEE (Main) Mathematics – Permutations & Combinations
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