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Q2:
How many 4-digit numbers are there with no digit repeated?
Solution :
Step 1: Identifying the Constraints
To form a 4-digit number using the set of available digits $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, we must satisfy the following conditions:
- The number consists of 4 distinct positions: Thousands ($T$), Hundreds ($H$), Tens ($t$), and Units ($U$).
- No digit can be repeated [Constraint: Selection without replacement].
- The Thousands place ($T$) cannot be occupied by the digit $0$, as a number starting with $0$ is considered a 3-digit number.
Step 2: Calculating Possibilities for Each Position
We analyze the number of choices for each slot sequentially:
| Position | Constraint | Number of Choices |
|---|---|---|
| Thousands ($T$) | Cannot be $0$ | $9$ (Digits $1$ through $9$) |
| Hundreds ($H$) | Cannot be digit used in $T$, but can be $0$ | $9$ (Total $10$ digits - $1$ used in $T$) |
| Tens ($t$) | Cannot be digits used in $T$ or $H$ | $8$ (Total $10$ digits - $2$ already used) |
| Units ($U$) | Cannot be digits used in $T, H, \text{ or } t$ | $7$ (Total $10$ digits - $3$ already used) |
Step 3: Applying the Fundamental Principle of Counting
According to the Fundamental Principle of Multiplication, if one task can be performed in $m$ ways and a second task in $n$ ways, the sequence of tasks can be performed in $m \times n$ ways [Fundamental Counting Principle].
The total number of 4-digit numbers $N$ is calculated as:
$N = (\text{Choices for } T) \times (\text{Choices for } H) \times (\text{Choices for } t) \times (\text{Choices for } U)$
$N = 9 \times 9 \times 8 \times 7$
Step 4: Final Computation
$9 \times 9 = 81$
$81 \times 8 = 648$
$648 \times 7 = 4536$
Final Solution: The total number of 4-digit numbers with no digit repeated is $4536$.
More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.3
- Q1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
- Q3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
- Q4: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
- Q5: From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
- Q6: Find $n$ if ${}^{n–1}P_3 : {}^nP_4 = 1 : 9$.
- Q7(i): Find $r$ if (i) ${}^5P_r = 2 \cdot {}^6P_{r-1}$
- Q7(ii): Find $r$ if (ii) ${}^5P_r = {}^6P_{r-1}$.
- Q8: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
- Q9(i): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time,
- Q9(ii): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (ii) all letters are used at a time,
- Q9(iii): How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (iii) all letters are used but first letter is a vowel?
- Q10: In how many of the distinct permutations of the letters in MISSISSIPPI do the four $I$'s not come together?
- Q11(i): In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S,
- Q11(ii): In how many ways can the letters of the word PERMUTATIONS be arranged if the (ii) vowels are all together,
- Q11(iii): In how many ways can the letters of the word PERMUTATIONS be arranged if the (iii) there are always 4 letters between P and S?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
Chapters in CBSE - Class 11 Mathematics
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