Wiley Edge Permutations and Combination Quiz-2

Explanation:
Total Number of letters = 5
Therefore Number of permutations = 5!
=> 120

Explanation:
6 artists around the table can be seated in 6! ways.
The available spaces between the seated artists to place the 4 musicians. Since the table is circular, we have 6 spaces between the artists.
We need to choose 4 spaces for the musicians from these 6 spaces.
The number of ways to choose 4 spaces out of 6 is given by the combination C(6, 4) = 15.
Once the spaces for the musicians are chosen, arrange the musicians within those spaces in 4! ways.
Total arrangements = 6! * C(6, 4) * 4!

Explanation:
Total Number of letters = 5
Therefore Number of permutations = 5!
=> 120

Explanation:
Consider the group of 3 specific books as one unit.
Now 6 entities have to be arranged: the group of 3 books, along with the remaining 5 individual books.
The 6 entities can be arranged in 6! ways.
However, within the group of 3 books, the books can be arranged among themselves in 3! ways.
Total arrangements = 6! * 3!
Total arrangements = 720 * 6 = 4,320.

Explanation:
The word "EXAMINATION" has 6 vowels (E, A, I, O) and 6 consonants (X, M, N, T).
To arrange the consonants, we have 6! Ways.
We have 7 spaces (including the ends) where the vowels can be placed. Since no two vowels can be adjacent, we need to choose 6 spaces for the 6 vowels.
The number of ways to choose 6 spaces out of 7 is given by 7C6 = 7.
Within each space, the vowels can be arranged in 6! ways.
Therefore, the total number of arrangements is 6! * 7 * 6!

Explanation:
Number of ways in which boys can be seated = 6!
Now there are 7 spaces between the boys and all girls can be seated in those spaces such that boys and girl must be alternate
Number of ways in which girls can be seated = C(7,6)* 6! = 7!
Total arrangements = 6! * 7!

Explanation:
Since the even digits need to occupy even positions, start by arranging the even digits (2, 4, 6) among themselves. There are 3! ways to arrange these even digits.
Similarly,There are 4! ways to arrange odd digits at odd places.
Therefore,Total number of arrangements = 3! * 4!

Explanation:
Since repetitions are allowed and there are 26 letters in the English alphabet, each letter in the password can be chosen in 26 ways.
Therefore, the total number of different passwords is 26^5

Explanation:
In the word "INDEPENDENCE", we have 5 vowels (I, E, E, E, E) and 7 consonants (N, D, P, N, D, N, .
To arrange the consonants, we have 7! ways.
Now, let's consider the spaces between the consonants. There are 9 spaces where the vowels can be placed. Since no two vowels can be adjacent, we need to choose 5 spaces for the 5 vowels.
The number of ways to choose 5 spaces out of 9 is given by 9C5 = 9! / (5! * 4!)
Within each space, the vowels can be arranged in 5! Ways.
Number of times E is appearing = 4
Number of Ways vowels can be arranged = 9! * 5! / (5! * 4!) = 9!/4!
Number of times N is appearing = 3
Number of times D is appearing = 2
Therefore, the total number of arrangements is 7! * 9! * 5! / (3! * 2! * 4! * 4!)

Explanation:
Since the 4 friends must always stand together, treat them as a single unit.
This reduces the problem to arranging 7 objects, where one of the objects is the group of 4 friends.
The number of ways to arrange 7 objects is 7!. However, within the group of 4 friends, they can be arranged among themselves in 4! ways.
Total number of arrangements = 7! * 4!