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 = \( (4 \times 3) \times 3 \times 3 \times 3 = 324 ) \)
How many positive divisors of \( 2^55^3{11}^4 \) are perfect squares?
60
18
120
4
Key Concepts
Basic counting principles
Divisors of a number
Check the Answer
Answer: is 18
ISI MStat 2019 PSA Problem 11
A First Course in Probability by Sheldon Ross
Try with Hints
See in order to get the positive divisors of \( 2^55^3{11}^4 \) that are perfect squares , we need to take only the even powers of the primes {2,5,11}.
Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .
The maximum powers of 5 is 3 . So there are 2 choices for even powers {0,2}.
Now the maximum powers of 2 is 5 . So there are 3 choices for even powers {0,2,4} .
The maximum powers of 11 is 4 . So there are 3 choices for even powers {0,2,4}.
Hence by multiplication principle we have in total \( 3 \times 2 \times 3 =18 \) such positive divisors.
