Question

How many numbers between 3000 and 4000 can be formed from the digits 3, 4, 5, 6, 7, and 8 with no digits being repeated in any number?
(a) 20
(b) 15
(c) 60
(d) 120

Answer 1

Hint:
Here, we will use the fact that all the numbers lying between 3000 and 4000 will have the digit 3 at the thousands place. We will fix 3 at the thousands place. Then, we will find the number of ways in which the remaining 5 digits can be placed in the hundreds, tens, and units place, to get the required number of ways.

Complete step by step solution:
The given digits are 3, 4, 5, 6, 7, 8. If the number starts with 4, it will lie between 4000 and 5000. Similarly, if the number starts with 5, 6, 7, 8, it will lie between 5000 and 6000, 6000 and 7000, 7000 and 8000, 8000 and 9000 respectively.
Therefore, all the numbers lying between 3000 and 4000 will have the digit 3 at the thousands place.
We will fix the digit 3 at the thousands place.
Therefore, the number of ways to place a digit at the thousands place such that the number lies between 3000 and 4000 is 1.
The remaining digits are 4, 5, 6, 7, and 8.
We can place any of the remaining 5 digits in the hundreds place.
Therefore, the number of ways to place a digit at the hundreds place such that the number lies between 3000 and 4000 is 5.
Now, 3 is placed in the thousands place, and a second digit is placed in the hundreds place.
We can place any of the remaining 4 digits in the tens place.
Therefore, the number of ways to place a digit at the tens place such that the number lies between 3000 and 4000 is 4.
Next, 3 is placed in the thousands place, a second digit is placed in the hundreds place, and a third digit is placed at the tens place.
We can place any of the remaining 3 digits in the units place.
Therefore, the number of ways to place a digit at the units place such that the number lies between 3000 and 4000 is 3.
Finally, we will calculate the number of ways in which a number between 3000 and 4000 can be formed from the digits 3, 4, 5, 6, 7, and 8.
The required number of ways is the product of the number of ways of placing a digit in the thousands place, hundreds place, tens place, and units place respectively.
Therefore, we get the number of ways to form a number between 3000 and 4000 from the digits 3, 4, 5, 6, 7, and 8, as \[1 \times 5 \times 4 \times 3 = 60\] ways.
Thus, there are 60 numbers lying between 3000 and 4000 that can be formed using the digits 3, 4, 5, 6, 7, and 8, where no digit is repeated.

Therefore, the correct option is option (c).

Note:
We can use the formula for permutations to find the required number of ways.
All the numbers lying between 3000 and 4000 will have the digit 3 at the thousands place.
We will fix the digit 3 at the thousands place.
The remaining digits are 4, 5, 6, 7, and 8.
We can place any of the remaining 5 digits in the remaining 3 places.
The number of permutations of a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\].
The number of ways to place any of the remaining 5 digits in the remaining 3 places is given by \[{}^5{P_3}\].
Therefore, substituting \[n = 5\] and \[r = 3\] in the formula, we get
\[ \Rightarrow {}^5{P_3} = \dfrac{{5!}}{{\left( {5 - 3} \right)!}} = \dfrac{{5!}}{{2!}}\]
Simplifying the expression, we get
\[ \Rightarrow {}^5{P_3} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 60\]