how many times does the digit 3 occurs at tens place in natural number from 100
Answers
Step-by-step explanation:
Let’s suppose XY3 is a 3 digit number with 3 in unit’s place
Number of ways X can be filled = 9 (1–9 numbers excluding 0 as it is a 3 digit number).
Number of ways Y can be filled = 10 (0–9 numbers)
Total number of numbers with 3 in unit’s place = 9*10 = 90
Step2 -
Now, let’s suppose X3Y is a 3 digit number with 3 in ten’s place.
Number of ways X can be filled = 9 (1–9 numbers excluding 0 as it is a 3 digit number).
Number of ways Y can be filled = 9 (0–9 numbers excluding 3 as we are counting this in Step1)
Total number of numbers with 3 in ten’s place = 9*9 = 81
So, number of times 3 occurs in ten’s and unit’s place from 100 to 1000 = 90+81 = 171.
The given question is we have to find how many times does the digit 3 occurs at tens place in natural number from 100
Lets solve this problem by the following steps.
let X3Y is a 3 digit number with 3 in ten's place.
The no of ways the x can be filled with digits =(1-9)
se excluded the 0 here, because 0 in the first place makes it invalid
The no of ways the Y can be filled with digits= (0-9) =10.
The no of times 3 occurs in 10 th place is 9*10=90
It is from 100 to 1000 only.
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