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?
Hint: From $0$ to $9$, two digits are already fixed for first and second position. We have to fill the remaining three positions with the remaining 8 digits.
According to the question, $5$ digit telephone numbers are to be formed using digits $0$ to $9$ keeping in mind that no digit is repeated. And the number must start with $67$. So out of the five places in the number, first two places are already taken by $67$. So we are left with only $3$ other places which are to be filled. Further, from $0$ to $9$ there are $10$ digits. Out of them, digit $6$ and digit $7$ is already taken for first and second place respectively. So, now we are left with only $8$ more digits (because we have to avoid repetition of digits also). Let’s suppose we are filling third place first. So, we have $8$ different digits for this place to be filled with and this can be done in $8$ different ways. $\therefore $ The number of ways of filling third place is $8$. Now, we are left with only $7$ more digits. So, the fourth place can be filled in $7$ different ways. $\therefore $ The number of ways of filling fourth place is $7$. For the last place, we have $6$ remaining digits. So, the last place can be filled in $6$ different ways. $\therefore $ The number of ways of filling last place is $6$. Therefore, by multiplication principle, the required number of ways in which five digit telephone numbers can be constructed is $8 \times 7 \times 6 = 336$.
Note: According to multiplication principle, if one event can occur in $m$ ways and a second event can occur in $n$ ways after the first event has occurred, then the two events can occur in $m \times n$ ways. This is also known as the Fundamental Counting Principle.
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