The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = 7C3 = 35

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

Preemptive scheduling: The preemptive scheduling is prioritized. The highest priority process should always be the process that is currently utilized.

Non-Preemptive scheduling: When a process enters the state of running, the state of that process is not deleted from the scheduler until it finishes its service time.

P(first letter is not vowel) = 24

P(second letter is not vowel) = 13

So, probability that none of letters would be vowels is = 24×13=16

Routable protocols can work with a router and can be used to build large networks. Non-Routable protocols are designed to work on small, local networks and cannot be used with a router

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.