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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

Check the Answer


Answer: is 324

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 = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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

Check the Answer


Answer: is 324

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 = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)

Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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