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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 \( \{1,2,…,1000\} \)?

- 590
- 600
- 300
- 299

Basic counting principles

But try the problem first...

Answer: is 300

Source

Suggested Reading

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

Content

[hide]

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 \( \{1,2,…,1000\} \)?

- 590
- 600
- 300
- 299

Basic counting principles

But try the problem first...

Answer: is 300

Source

Suggested Reading

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

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