Question:
How many words (with or without meaning) can be formed using vowels? (Repetition allowed and not allowed)
Vowels are: A, E, I, O, U → 5 vowels.
Answer:
If repetition is allowed:
Each position can have any vowel.
So,
5×5×5=53=125 words
If repetition is not allowed:
Each position must have a different vowel.
So,
5×4×3=60 words
Explanation (Simple Language):
Since all letters are used together, it’s a case of multiplication.
With repetition → Same letter can appear again.
Without repetition → No letter can be repeated; choices decrease.
Question:
There are 4 letters and 5 letterboxes. In how many ways can the letters be posted?
Answer:
Repetition is allowed — because a letterbox can receive multiple letters.
(No rule that one box can hold only one letter.)
So,
5×5×5×5=54 =625 ways
Short Approach Explanation:
Letters are moving (not letterboxes).
Make blanks for moving items → (Letter1, Letter2, Letter3, Letter4).
Fill each blank with number of options (letterboxes = 5).
⚡ Note:
Answer is not 45 because letterboxes cannot move toward the letters — that would be weird!
Question:
There are 3 rings and 4 fingers. In how many ways can the rings be worn?
Answer:
No restriction (repetition allowed):
Rings are moving to fingers.
So,
4×4×4=43 =64 ways
If repetition is not allowed (one finger holds only one ring):
4×3×2=24 ways
Short Approach Explanation:
Rings are movable → make blanks for rings.
Fill each blank with number of fingers (4 options initially).
🚫 Wrong Approach:
If you mistakenly do 34, it means fingers are moving to rings — not natural during a wedding ceremony (and sounds like a forced marriage 😅).
Question:
There are 5 persons and 8 chairs. How many seating arrangements are possible?
Answer:
Question:
There are 8 persons and 5 chairs. How many seating arrangements are possible?
Answer:
This question is senseless in basic counting because you can’t seat 8 persons on 5 chairs without extra conditions like:
Sitting on laps (which is not usual in standard problems 😅)
Or standing/switching chairs (not considered here)
Thus, invalid without additional context.
Question:
How many 3-letter passwords can be formed using 6 different alphabets if repetition is allowed?
Answer:
Each position has 6 choices.
6×6×6=63=216 passwords
Question:
A captain wants to select either a bowler (6 choices) or a batsman (5 choices) for a match. How many ways can he select?
Answer:
Feeling of either/or → Addition principle.
6+5=11 ways
Practicing these basic counting problems builds a strong foundation for more advanced topics like permutations, combinations, and probability.
Always remember:
Multiplication → Feeling of “and”, “all”, “together”.
Addition → Feeling of “or”, “either”, “only one”.
Next Up: We’ll explore permutations in detail with real-world examples!