Mẹo How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
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How many numbers are there between $ 100 $ and $ 1000 $ such that least one of their digits is $ 7 $ ?
Let's count the number of occurrences with only one 7:
(1) 7XX: 1 * 9 * 9 = 81 (2) X7X: 8 * 1 * 9 = 72 (3) XX7: 8 * 9 * 1 = 72 81 + 72 + 72 = 225 At this point we can already see that the answer must be D since the choice are very dispersed. Anyways, let's count number of occurrences with two 7s: (1) 77X: 1 * 1 * 9 = 9 (2) X77: 8 * 1 * 1 = 8 (3) 7X7: 1 * 9 * 1 = 9 9 + 8 + 9 = 26 Number
777 is missing. So the total becomes
225 + 26 + 1 = 252 How many numbers are between 100 and 1000 which have exactly one of their digits as 5?How many numbers from 1 to 1000 contains the digit 7?How many numbers are there between 100 and 1000 which have exactly one of their digits as 6?How many numbers are there between 100 and 1000 which have exactly one of their digits as 8?
How many numbers are there between $ 100 $ and $ 1000 $ such that least one of their digits is $ 7 $ ?
Answer
Verified
Hint:
For this problem, we will use the concept of permutation and combinations. By using this concept, we will calculate the number of possible ways to make a $ 3 $ digit number which lies between $ 100 $ and $ 1000 $ such that least one of their digits is $ 7 $. For this, we will consider three cases. The first one is “the number having only one $ 7 $ ”, the second one is “the number having two $ 7 $ ’s”, and the third case is “the number having all three digits as $ 7 $ ”.
We will add all of them to get the result.
Complete step by step answer:
Assuming a three digited number, which lies between $ 100 $ and $ 1000 $.
Case-1: Only one digit is $ 7 $ .
(i) When the one’s place is $ 7 $ .
The one’s place can be arranged in one way, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways to get a number greater than $ 100 $ we neglected $ 0 $ in the tens place, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a
number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1times 9times 8=72 $ .
(ii) When the tens place is $ 7 $ .
The one’s place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens place is only filled by $ 7 $ in one way, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1times
9times 8=72 $ .
(iii) When the hundreds place is $ 7 $ .
The ones place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways to get a number greater than $ 100 $ we neglected $ 0 $ in tens place, hundreds place can be filled by $ 7 $ in one way.
Hence the total number of ways is $ 1times 9times 9=81 $ .
Case-2: When two digits are $ 7 $ .
(i) when ones and tens place are filled with $ 7 $ .
Ones and tens
place are filled with $ 7 $ in only one way, hundreds place can be filled by $ (2,3,4,5,6,7,8,9) $ in $ 8 $ ways to get a number greater than $ 100 $ , we neglected $ 0,1 $ in the hundreds place.
Hence the total number of ways is $ 1times 8=8 $ .
(ii) When hundreds and tens place is $ 7 $ .
The one’s place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways, tens and hundreds place can be filled by $ 7 $ in only one way.
Hence the total number of ways is $ 1times 9=9 $
.
(iii) When ones and hundred place is $ 7 $ .
The ones and hundreds place can be filled by $ 7 $ in one way, tens place can be filled by $ (1,2,3,4,5,6,7,8,9) $ in $ 9 $ ways.
Hence the total number of ways is $ 1times 9=9 $ .
Case-3: When all three digits are $ 7 $ .
We have only one possible way to fill the $ 7 $ in all three digits.
Hence the total number of ways is $ 1 $ .
Now all the possible ways to form a digit which is having least one digit as $ 7 $ between $
100 $ and $ 1000 $ is $ 72+72+81+8+9+9+1=252 $ .
Note:
We can simply solve this problem by calculating the number of possible ways to form a number which do not have $ 7 $ as a digit between $ 100 $ and $ 1000 $. We can fill ones, tens place in $ 9 $ ways by $ (1,2,3,4,5,6,7,8,9) $ and hundred places can be filled in $ 8 $ ways by $ (2,3,4,5,6,7,8,9) $ , hence the possible ways are $ 8times 9times 9=648 $ . Now the required value can be obtained by subtracting the
calculated value from a number of numbers between $ 100 $ and $ 1000 $. We have $ 1000-100=900 $ numbers between $ 100 $ and $ 1000 $ .
$ beginalign
& therefore x=900-648 \
& Rightarrow x=252 \
endalign $
From both the methods we got the same result.
Solution
Let's count the number of occurrences with only one 7: (1) 7XX: 1 * 9 * 9 = 81 (2) X7X: 8 * 1 * 9 = 72 (3) XX7: 8 * 9 * 1 = 72 81 + 72 + 72 = 225 At this point we can already see that the answer must be D since the choice are very dispersed. Anyways, let's count number of occurrences with two 7s: (1) 77X: 1 * 1 * 9 = 9 (2) X77: 8 * 1 * 1 = 8 (3) 7X7: 1 * 9 * 1 = 9 9 + 8 + 9 = 26 Number 777 is missing. So the total becomes 225 + 26 + 1 = 252
Solution : Clearly, the numbers between 100 and 1000 are 3-digits numbers.
So, we have to form 3-digit numbers having exactly one of their digits as 7.
Case I 3-digit numbers having 7 unit's place only:
Number of ways to fill the unit's place `=1 " " `[ with 7 only].
Number of ways to fill the ten's place `=9 " " `[any digit from 0 to 9 except 7].
Number of ways to fill the hundred's place `=8 " " `[
any digit from 1 to 9 except 7].
So, the number of such numbers `=(1xx9xx8)=72.`
Case II 3-digit numbers having 7 ten's place only :
Number of ways to fill the ten's place `=1 " " `[with 7 only].
Number of ways to fill the unit's place `= 9 " " `[any digit from 0 to 9 except 7].
Number of ways to fill the hundred's place `=8 " " `[any digit from 1 to 9 except 7].
So, the number of such numbers `=(1xx9xx8)=72.`
Case
III 3 - digit numbers having 7 hundred's place only:
Number of ways to fill the hundred's place`=1 " " `[with 7 only].
Number of ways to fill the ten's place `=9 " " `[ any digit from 0 to 9 except 7].
Number of ways to fill the unit's place `= 9 " " `[ any digit from 0 to 9 except 7].
So, the number of such numbers `= (1xx9xx9) =81`.
Hence, the total number of required numbers `=(72+72+81)=225.`
How many numbers are between 100 and 1000 which have exactly one of their digits as 5?
The number of numbers between 100 and 1000 such that exactly one of the digits is 5 (with repetition) is: A) 196.How many numbers from 1 to 1000 contains the digit 7?
∴ The number of times 7 will be written when listing the numbers from 1 to 1000 is 300.How many numbers are there between 100 and 1000 which have exactly one of their digits as 6?
Thus, there are 225 numbers between 100 and 1000 that have exactly one of the digits as 6.How many numbers are there between 100 and 1000 which have exactly one of their digits as 8?
Number of ways to fill the ten's place `=9 " " `[any digit from 0 to 9 except 7].
Number of ways to fill the hundred's place `=8 " " `[ any digit from 1 to 9 except 7].
So, the number of such numbers `=(1xx9xx8)=72. Tải thêm tài liệu liên quan đến nội dung bài viết How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
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