Question
Question: The number of times the digit 5 will be written when listing the integers from 1 to 1000 is A.271 ...
The number of times the digit 5 will be written when listing the integers from 1 to 1000 is
A.271
B.272
C.300
D.None of these
Solution
Here, we will consider three different cases i.e. when the digit 5 appears only once, when it appears twice and when it appears thrice using the formula of combination. We will then add all these 3 cases to get the required number of times.
Formula Used:
nCr=r!(n−r)!n!, where n is the total number of terms and r is the number of terms to be selected.
Complete step-by-step answer:
We are going to find the number of times the digit 5 appears while listing the numbers from 1 to 1000.
Now, while listing the numbers from 1 to 1000, we can have a maximum of 3 digit numbers.
Hence, we will consider 3 cases:
Case 1: When the digit 5 appears only once.
Now, as we know that we can have a maximum of three digit numbers.
Hence, in order to choose only 1 place from those 3 digits, we will use the formula of combinations nCr
Hence, the number of ways of choosing that 1 place where 5 will appear =3C1
Also, the remaining 2 digits can be filled by the remaining 9 digits from 0 to 9 except 5.
The total number of ways of filling the remaining two digits =9×9=92
The total number of cases where the digit 5 appears only once =3C1×(9)2
Using the formula, nCr=r!(n−r)!n!, we get,
Total number of cases where the digit 5 appears only once =1!(2)!3!×(9)2
Simplifying the expression, we get
⇒ Total number of cases where the digit 5 appears only once =3×81=243
Case 2: When the digit 5 appears twice.
Now, as we know that we can have a maximum of three digit numbers.
Hence, in order to choose 2 places from those 3 digits, we will use the formula of combinations nCr
Hence, the number of ways of choosing 2 places where 5 will appear =3C2
Also, the remaining 1 digit can be filled by the remaining 9 digits from 0 to 9 except 5.
Hence, total number of ways of filling the remaining digit will be 9
Also, the digit 5 appears twice, hence, we will multiply by 2
Hence, total number of cases where the digit 5 appears twice =2×3C2×9
Using the formula, nCr=r!(n−r)!n!, we get,
Total number of cases where the digit 5 appears two times =2×2!(1)!3!×9
Simplifying the expression, we get
⇒ Total number of cases where the digit 5 appears two times =2×3×9=54
Case 3: When the digit 5 appears three times
Now, as we know that we can have a maximum of three digit numbers.
Hence, this case will include only 1 possible case which is having a number 555.
But since, we have the same number 5 three times, hence, we will multiply this by 3.
Hence, total number of possible cases =1×3=3
Therefore, adding all the three cases together, we get
Required number of times the digit 5 occurs =243+54+3=300
Therefore, the digit 5 occurs 300 times.
Hence, option C is the correct answer.
Note: We can solve this question in an alternate way.
First, we will think that how many 5’s are at the units place:
\left\\{ {5,15,25,35,....95} \right\\} = 10 fives
Then how many 5’s are at the tens place:
\left\\{ {50,51,52,53,....59} \right\\} = 10 fives
Hence, from 1 to 100, we have a total of 10+10=20 fives.
Hence, from 1 to 1000, we will have 20×10=200 fives in the units and tens place altogether.
Therefore, at the hundred’s place, the number of 5’s are:
\left\\{ {500,501,502,.....599} \right\\} = 100 fives
Hence, the total number of times the digit 5 will be written when listing the integers from 1 to 1000 is 200+100=300 times.
Hence, option C is the correct answer.