This is a beautiful problem from ISI MSTAT 2019 PSA problem 6 based on basic counting principles. We provide sequential hints so that you can try.
How many times does the digit ‘2’ appear in the set of integers \( \{1,2,…,1000\} \)?
Basic counting principles
But try the problem first...
Answer: is 300
ISI MStat 2019 PSA Problem 6
A First Course in Probability by Sheldon Ross
First hint
Let’s count the number of times 2 occurs once.
_ _ _
The position of 2 can be selected in 3 ways. The rest in 9 x 9.
Total number of 2 = 1 x 3 x 9 x 9 = 243
Second Hint
Let’s count the number of times 2 occurs twice.
_ _ _
The positions of 2 can be selected in 3 ways. The rest in 9.
Total number of 2 = 2 x 3 x 9 = 54
Final Step
Let’s count the number of times 2 occurs thrice.
The positions of 2 can be selected in 1 way. The rest in 1.
Total number of 2 = 3 x 1 x 1 = 3
Total Number of Such Numbers = 243 + 54 + 3 = 300
This is a beautiful problem from ISI MSTAT 2019 PSA problem 6 based on basic counting principles. We provide sequential hints so that you can try.
How many times does the digit ‘2’ appear in the set of integers \( \{1,2,…,1000\} \)?
Basic counting principles
But try the problem first...
Answer: is 300
ISI MStat 2019 PSA Problem 6
A First Course in Probability by Sheldon Ross
First hint
Let’s count the number of times 2 occurs once.
_ _ _
The position of 2 can be selected in 3 ways. The rest in 9 x 9.
Total number of 2 = 1 x 3 x 9 x 9 = 243
Second Hint
Let’s count the number of times 2 occurs twice.
_ _ _
The positions of 2 can be selected in 3 ways. The rest in 9.
Total number of 2 = 2 x 3 x 9 = 54
Final Step
Let’s count the number of times 2 occurs thrice.
The positions of 2 can be selected in 1 way. The rest in 1.
Total number of 2 = 3 x 1 x 1 = 3
Total Number of Such Numbers = 243 + 54 + 3 = 300
