Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed ?
6
For numbers greater than 30000, 2 will not be placed as the first digit. So there are 4 places to fill with 2, 2, 3, 3.
Distinct numbers greater than 30000 formed = \(\frac{4!}{2!\times 2!}\)
= \(\frac{24}{2\times 2}\) = 6
Hence, option 2 is correct.
Important Points
The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, is \(\frac{n!}{p!q!r!}\)
Important Questions from Integers
How many three digit whole numbers are there between 75 and 405?
Find the number of integers between $1$ and $150$ (inclusive) having $7$ as one of the digits but which are not divisible by $7$.
Integers are listed from 700 to 1000. In how many integers is the sum of the digits 10 ?
Consider the following statements :
1. The sum of 5 consecutive integers can be 100.
2 The product of three consecutive natural numbers can be equal to their sum.
Which of the above statements is/are correct ?
The difference between a 2-digit number and the number obtained by interchanging the positions of the digits is 54.
Consider the following statements:
1. The sum of the two digits of the number can be determined only if the product of the two digits is known.
2. The difference between the two digits of the number can be determined.
Which of the above statements is/are correct?
