Step 1: Understand the problem.
We need to form words using 3 consonants and 3 vowels, chosen from 5 consonants and 4 vowels.
Step 2: Calculate the number of ways to choose 3 consonants and 3 vowels.
- The number of ways to choose 3 consonants from 5:
\[
\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
\]
- The number of ways to choose 3 vowels from 4:
\[
\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4
\]
Step 3: Multiply the results and consider arrangements.
Now, the 6 letters (3 consonants and 3 vowels) can be arranged in:
\[
\frac{6!}{3!3!} = \frac{720}{6 \times 6} = 20
\]
Thus, the total number of words that can be made is:
\[
10 \times 4 \times 20 = 80
\]
Therefore, the correct answer is 2. 80.
In C language, mat[i][j] is equivalent to: (where mat[i][j] is a two-dimensional array)
Suppose a minimum spanning tree is to be generated for a graph whose edge weights are given below. Identify the graph which represents a valid minimum spanning tree?
\[\begin{array}{|c|c|}\hline \text{Edges through Vertex points} & \text{Weight of the corresponding Edge} \\ \hline (1,2) & 11 \\ \hline (3,6) & 14 \\ \hline (4,6) & 21 \\ \hline (2,6) & 24 \\ \hline (1,4) & 31 \\ \hline (3,5) & 36 \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match LIST-I with LIST-II
Choose the correct answer from the options given below:
Consider the following set of processes, assumed to have arrived at time 0 in the order P1, P2, P3, P4, and P5, with the given length of the CPU burst (in milliseconds) and their priority:
\[\begin{array}{|c|c|c|}\hline \text{Process} & \text{Burst Time (ms)} & \text{Priority} \\ \hline \text{P1} & 10 & 3 \\ \hline \text{P2} & 1 & 1 \\ \hline \text{P3} & 4 & 4 \\ \hline \text{P4} & 1 & 2 \\ \hline \text{P5} & 5 & 5 \\ \hline \end{array}\]
Using priority scheduling (where priority 1 denotes the highest priority and priority 5 denotes the lowest priority), find the average waiting time.