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.
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?
Basic counting principles
But try the problem first...
Answer: is 324
ISI MStat 2018 PSA Problem 14
A First Course in Probability by Sheldon Ross
First hint
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.
\( 4 \times 3 \).
Second Hint
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.
Final Step
Therefore, total number of ways to colour = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)
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.
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?
Basic counting principles
But try the problem first...
Answer: is 324
ISI MStat 2018 PSA Problem 14
A First Course in Probability by Sheldon Ross
First hint
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.
\( 4 \times 3 \).
Second Hint
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.
Final Step
Therefore, total number of ways to colour = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)
