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Q4:
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Solution :
Given Variables & Initial Setup
A 5-digit telephone number is to be constructed using the set of available digits $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The total number of available digits $n = 10$.
- Total positions available: $5$
- Fixed starting digits: $67$ (occupying the first two positions).
- Constraint 1: The first two positions are occupied by $6$ and $7$.
- Constraint 2: No digit can be repeated (sampling without replacement).
Step 1: Identifying Remaining Positions and Digits
Since the first two positions are fixed as $67$, we are required to fill the remaining $3$ positions ($5 - 2 = 3$).
The set of available digits for the remaining positions is $S' = S \setminus \{6, 7\} = \{0, 1, 2, 3, 4, 5, 8, 9\}$.
The number of available digits remaining is $n' = 10 - 2 = 8$.
Step 2: Applying the Fundamental Principle of Counting
We must select and arrange $3$ digits from the set of $8$ available digits to fill the $3$ vacant spots. Per the [Fundamental Counting Principle], if an event can occur in $m$ ways and a subsequent event in $n$ ways, the total occurrences are $m \times n$.
| Position | Available Choices |
|---|---|
| 3rd Digit | $8$ options |
| 4th Digit | $7$ options (since one digit was used for the 3rd position) |
| 5th Digit | $6$ options (since two digits were used for previous positions) |
Step 3: Calculating Permutations
The number of ways to arrange $r$ objects from $n$ distinct objects is given by the formula $P(n, r) = \frac{n!}{(n-r)!}$.
Here, $n = 8$ and $r = 3$:
$P(8, 3) = \frac{8!}{(8-3)!}$
$P(8, 3) = \frac{8 \times 7 \times 6 \times 5!}{5!}$ [By the definition of factorial simplification]
$P(8, 3) = 8 \times 7 \times 6 = 336$
Final Solution: The total number of 5-digit telephone numbers that can be constructed is 336.
More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.1
- Q1(i): How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed?
- Q1(ii): How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (ii) repetition of the digits is not allowed?
- Q2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
- Q3: How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
- Q5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
- Q6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
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