Question Number 77777 by jagoll last updated on 10/Jan/20

Answered by mr W last updated on 10/Jan/20

Commented by jagoll last updated on 10/Jan/20

Commented by mr W last updated on 10/Jan/20

Commented by jagoll last updated on 10/Jan/20

Commented by mr W last updated on 10/Jan/20

Commented by jagoll last updated on 10/Jan/20

Commented by jagoll last updated on 10/Jan/20

Commented by malwaan last updated on 10/Jan/20

Commented by jagoll last updated on 10/Jan/20

Commented by mr W last updated on 10/Jan/20
![to jagoll sir: image you mark the three letters A with a pen as A_1 ,A_2 ,A_3 and treat them as different letters. certainly they are in fact identical. then we select three vowols, e.g. A_1 ,I,U, as a group, say [A_1 IU], as if they were a single letter. then we arrange the 6 “letters”: [A_1 IU],A_2 ,A_3 ,M,S,T. there are 6! ways to arrange them. but A_2 and A_3 are in fact identical, therefore there are in fact only ((6!)/(2!)) different ways. on the other side, we can arrange the three vowols A_1 ,I,U inside the group in 3! ways. therefore we have totally ((6!3!)/(2!)) ways when we take A_1 as vowol. since we can also select A_2 or A_3 as vowol, the number of all possibilties is ((3×6!3!)/(2!)). clear?](https://www.tinkutara.com/question/Q77797.png)
Commented by jagoll last updated on 10/Jan/20

Commented by malwaan last updated on 10/Jan/20
