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# ISI MStat PSA 2019 Problem 6 | Basic Counting principles

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.

## Basic Counting Principles - ISI MStat Year 2019 PSA - 6

How many times does the digit ‘2’ appear in the set of integers $$\{1,2,…,1000\}$$?

• 590
• 600
• 300
• 299

### Key Concepts

Basic counting principles

## Check the Answer

ISI MStat 2019 PSA Problem 6

A First Course in Probability by Sheldon Ross

## Try with Hints

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

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

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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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.

## Basic Counting Principles - ISI MStat Year 2019 PSA - 6

How many times does the digit ‘2’ appear in the set of integers $$\{1,2,…,1000\}$$?

• 590
• 600
• 300
• 299

### Key Concepts

Basic counting principles

## Check the Answer

ISI MStat 2019 PSA Problem 6

A First Course in Probability by Sheldon Ross

## Try with Hints

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

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

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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