00:01
In this question we have the following letters given a, b, c, d, e and f.
00:07
So first calculating the number of permutations, when there are no restrictions, when there are no restrictions, the total number of permutations is calculated as total number of permutations is equal to 6 c6 p6.
00:31
This is nothing but 6 factorial.
00:35
Therefore 6 times of 5 times of 4 times of 3 times of 2 times of 1 therefore the total number of permutations when there are no restrictions is equal to 720 now next calculating we are given that the letters a and b are to be adjacent the letters a and b are to be adjacent to each other are to be adjacent to each other, adjacent to each other.
01:10
So first let us consider a and b together as a unit, then we will have the total number of letters is c, d, e and f.
01:19
So five letters since we are considering a and b as a unit.
01:23
Then we have a total of five letters.
01:25
So the number of permutations will be equal to, so the number of permutations is equal to 5 factorial and the letters a and b can be arranged or interchanged among themselves in two factorial ways so this is the required permutations now next we have on calculating this we get the required answer as 240 now next calculating so we are given the first letter is a the first letter is a b c and the last letter is d e f and the last letter, last letter is d, e and f.
02:12
So we can first fill first letter in three ways, which is a, b, c, and we can fill the last letter in three ways, which is d, e, f.
02:21
So the middle four places can be filled in four factorial ways, four factorial times, and the first three letters and the last three letters can be filled in three times of three ways.
02:32
So the total number of ways is equal to 216.
02:35
So this is the required solution for the third part.
02:39
Now next we have...