00:01
Hello everyone.
00:02
So in this question the question reads that how many different arrangements can be made of all the letters of the word accountants? a -c -c -o -u -n -t -a -n -t -a -n -t -s.
00:15
Also we have to tell in how many of them the o -w will stand together.
00:19
So coming to part a of the question, we see that there are total of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
00:28
Total of 11 letters of which there are repetitions of a, there are 2a, repetition of c, there are 2c, repetition of n, there are 2n and t, there are 2n, and t, there are 2t.
00:51
So, number of possible permutations become total possible permutation, becomes 11 factorial divided by repeated letters that are 2 factorial, 2 factorial and 2 factorial.
01:28
On solving this, we get this is equal to 11 factorial is equals to 399168 00 divided by 2 factorial multiplied by 2 factorial, multiplied by 2 factorial, multiplied by 2 factorial.
01:44
Now 2 factorial is equal to 2 multiplied by 1, that is 2.
01:52
So this becomes 16.
01:54
So on further solving, we get this is equals to 2499480, that is a little bit under 2 .5 medium.
02:07
Coming to part b of the question in which we have to tell the number of possible permitting when vowels are grouped together.
02:21
Vowels are grouped.
02:27
Now we have accountants...