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# ISI MStat 2018 PSA Problem 14 | All possible colorings

Try this problem from ISI MStat 2018 PSA Problem 14 based on all possible colorings. We provide sequential hints to help you solve the problem.

## Probability - ISI MStat 2018 PSA 14

A flag is to be designed with 5 vertical stripes using some or all of the four colours: green, maroon, red and yellow. In how many ways can this be done so that no two adjacent stripes have the same colour?

• 576
• 120
• 324
• 432

### Key Concepts

Basic counting principles

ISI MStat 2018 PSA Problem 14

A First Course in Probability by Sheldon Ross

## Try with Hints

How many ways you can colour the first stripe and the second stripe?

You have 4 colour options for the first stripe.
You have (4-1) = 3 colour options for the second stripe.
$4 \times 3$.

How many ways you can colour the third stripe and the rest of the stripes?

You have (4-1) = 3 colour options for the second stripe.
You have (4-1) = 3 colour options for the fourth stripe.
You have (4-1) = 3 colour options for the fifth stripe.

Therefore, total number of ways to colour = $(4 \times 3) \times 3 \times 3 \times 3 = 324 )$

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Try this problem from ISI MStat 2018 PSA Problem 14 based on all possible colorings. We provide sequential hints to help you solve the problem.

## Probability - ISI MStat 2018 PSA 14

A flag is to be designed with 5 vertical stripes using some or all of the four colours: green, maroon, red and yellow. In how many ways can this be done so that no two adjacent stripes have the same colour?

• 576
• 120
• 324
• 432

### Key Concepts

Basic counting principles

ISI MStat 2018 PSA Problem 14

A First Course in Probability by Sheldon Ross

## Try with Hints

How many ways you can colour the first stripe and the second stripe?

You have 4 colour options for the first stripe.
You have (4-1) = 3 colour options for the second stripe.
$4 \times 3$.

How many ways you can colour the third stripe and the rest of the stripes?

You have (4-1) = 3 colour options for the second stripe.
You have (4-1) = 3 colour options for the fourth stripe.
You have (4-1) = 3 colour options for the fifth stripe.

Therefore, total number of ways to colour = $(4 \times 3) \times 3 \times 3 \times 3 = 324 )$

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