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If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?
- a)3600
- b)3768
- c)3980
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If we listed all numbers from 100 to 10,000, how many times would the ...
Three-digit numbers: A B C. 3 can be printed in the 100’s place or 10’s place or units place.
=> 100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times
=> 10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times
=> Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times
So, 3 gets printed 280 times in 3-digit numbers
Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or 10’s place or units place.
=> 1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.
=> 100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.
=> 10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.
=> Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.
3 gets printed 3700 times in 4-digit numbers.
So, there are totally 3700 + 280 = 3980 numbers.
=> 100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times
=> 10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times
=> Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times
So, 3 gets printed 280 times in 3-digit numbers
Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or 10’s place or units place.
=> 1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.
=> 100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.
=> 10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.
=> Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.
3 gets printed 3700 times in 4-digit numbers.
So, there are totally 3700 + 280 = 3980 numbers.
Hence the answer is "3980", Choice C is the correct answer.
Most Upvoted Answer
If we listed all numbers from 100 to 10,000, how many times would the ...
There are 6 letters in the word 'LEADER'. To find the number of ways the letters can be arranged, we can use the concept of permutations.
Permutations refer to the arrangement of objects in a specific order. In this case, we want to find the number of ways the letters can be arranged.
Step 1: Identify the total number of objects
In this case, the total number of objects is 6, which represents the 6 letters in the word 'LEADER'.
Step 2: Determine if there are any repeating objects
In the given word 'LEADER', the letter 'E' appears twice. This means that we have repeating objects.
Step 3: Calculate the number of ways the objects can be arranged without considering the repeating objects.
Since we have 6 letters in total, we can arrange them in 6! (6 factorial) ways. The factorial of a number is the product of all positive integers less than or equal to that number. Therefore, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
Step 4: Adjust for the repeating objects
Since the letter 'E' appears twice, we need to adjust for the overcounting that occurs when we treat the two 'E's as distinct objects. To do this, we divide the total number of arrangements by the factorial of the number of repeating objects. In this case, we divide 720 by 2! (2 factorial) to account for the two 'E's. 2! = 2 x 1 = 2.
Step 5: Calculate the final number of arrangements
Dividing 720 by 2 gives us 360. Therefore, there are 360 ways to arrange the letters of the word 'LEADER'.
Therefore, the correct answer is option 'C' - 360.
Permutations refer to the arrangement of objects in a specific order. In this case, we want to find the number of ways the letters can be arranged.
Step 1: Identify the total number of objects
In this case, the total number of objects is 6, which represents the 6 letters in the word 'LEADER'.
Step 2: Determine if there are any repeating objects
In the given word 'LEADER', the letter 'E' appears twice. This means that we have repeating objects.
Step 3: Calculate the number of ways the objects can be arranged without considering the repeating objects.
Since we have 6 letters in total, we can arrange them in 6! (6 factorial) ways. The factorial of a number is the product of all positive integers less than or equal to that number. Therefore, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
Step 4: Adjust for the repeating objects
Since the letter 'E' appears twice, we need to adjust for the overcounting that occurs when we treat the two 'E's as distinct objects. To do this, we divide the total number of arrangements by the factorial of the number of repeating objects. In this case, we divide 720 by 2! (2 factorial) to account for the two 'E's. 2! = 2 x 1 = 2.
Step 5: Calculate the final number of arrangements
Dividing 720 by 2 gives us 360. Therefore, there are 360 ways to arrange the letters of the word 'LEADER'.
Therefore, the correct answer is option 'C' - 360.
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If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?a)3600b)3768c)3980d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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