Hướng Dẫn How many numbers are there between 100 and 1000 which have exactly one of their digits as 5?

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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 The number of numbers between 100 and 1000 such that exactly one of the digits is 5 (with repetition) is:A) 196B) 225C) 289D) 324 Related VideosHow many numbers are there between 100 and 1000 that have exactly one of their digits as 7?How many numbers are there between 100 and 1000 which have exactly one of their digits is 8?How many numbers are there between 100 and 1000 such that exactly one of the digits is 6?How many numbers are there between 100 and 1000 which have exactly one of the digits as 7 A 125 B 150 C 225 D 250?

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

How many numbers between 100 and 1000 have the digit 7 exactly once?

A number between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
Let us consider the three cases separately.
Case (I): the digit 7 is in the unit’s place.

The ten’s place is filled by one digit from 0 to excluding 7 in 9 ways.
Here, there are 8 × 9 × 1 = 72 three digit numbers with the required condition.
Case (II): The digit 7 is in the ten’s place.

Unit’s place can be filled by digit from 0 to 9 excluding 7 in 9 ways.
Zero is not allowed a hundred’s place.
Hundred's place can be filled by digit from 1 to 9 excluding 7 in 8 ways.
The hundred’s place can be filled in by any digit from 1 to 9 excluding 7 in 8 ways.
Here, there will be 8 × 1 × 9 = 72
3-digit numbers with the required condition
Case (III): The digit 7 is in the hundred’s place.

Then, there are 1 × 9 × 9 = 81
3-digit numbers with the required condition.
Hence, the numbers between 100 and 1000 having the digit 7 exactly once are 72 + 72 + 81 = 225.

The number of numbers between 100 and 1000 such that exactly one of the digits is 5 (with repetition) is:A) 196B) 225C) 289D) 324

Answer

Verified

Hint:
we know about the units’ place, tens place, and hundreds place. If we take units place as 5, then we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.

Complete step by step solution:
The number of numbers between 100 and 1000 such that exactly one of the digits is 5.
When the units place has 5, it means we can take tens place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 times 8 = 72$ numbers.
When the tens place has 5, it means we can take unit place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take hundreds place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled with 8 different ways.
Hence, there are $9 times 8 = 72$ numbers.
When the hundreds place has 5, it means we can take units place any one of the digits from 0 to 9 except 5. Therefore, it can be filled in 9 different ways. Now, we take tens place any one of the digits from 0 to 9 except 0 and 5. Therefore, it can be filled in 9 different ways.
Hence, there are $9 times 9 = 81$ numbers.
Now, calculate the required numbers of numbers is:
$72 + 72 + 81 = 225$

Hence, the number of numbers between 100 and 1000 is 225. The option (B) is the correct option.

Note:
Here you should know each digit has 10 choices from 0 to 9 for every range of numbers. The unit place, tens place and hundred place must be taken with the required conditions. Make sure to remember that the problem says with repetition.

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Updated On: 27-06-2022

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