How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM
A number between 100 and 1000 contains 3 digits. So, we have to form 3 digit numbers having exactly one of their digits as7. Such type of numbers can be divided into three types:
1) There numbers that have 7 in the unit place but not in any other place.
2) These numbers that have 7 in the ten's place but not in any other place.
3) These numbers that have 7 in the hundred's place but not in any other ways.
We shall now count these three type of numbers separately.
1) These 3-digit numbers that have 7 in the unit's place but not in any other place.
The hundred's place can have any one of the digits from 0 to 9 except 0 and 7. So hundred's place can be filled in 8 ways. The ten's place can have anyone of the digits from 0 to 9 except 7. So the number of ways the ten's place can be filled is 9. The unit's place has 7. So, it can be filled in only one way.
Thus, there are 8 x 9 x 1 = 72 numbers of the first kind.
2) Those three-digit numbers that have 7 in the ten's place but not in any other place.
The number of ways to fill the hundred's place = 8 [By any one of the digits from
The number of ways to fill the ten's place = 1 [By 7 only]
The number of ways to fill the one's place = 9 [By any one of the digits 0,1,2,3,4,5,
Thus, these are 8 x 1 x 9 = 72 numbers of the second kind.
3) These three digit numbers that have 7 in the hundred's place but not at any other place.
In this case, the hundred's place can be filled only in one way and each of the ten's and one's place can be filled in 9 ways.
So, there are 1 x 9 x 9 = 81 numbers of the third kind.
Hence, the total number of required type of numbers = 72 + 72 + 81 = 225.
Answered by | 4th Jun, 2014, 03:23: PM
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