There are 8 orators a, b, c, d, e, f, g and h. In how many ways can the arrangements be made so that a always comes before b and b always comes before
c.
Answers
Answered by
15
Answer:
6720
Step-by-step explanation:
There are 8 orators a, b, c, d, e, f, g and h. In how many ways can the arrangements be made so that a always comes before b and b always comes before
Total number of arrangements
= 8!
if a , b , c had no conditios then they can be arranged in
3! Ways
but a comes before b & b comes before c
=> only 1 Ways
Total Possible Combination = 8! * 1/3!
= 6720
Answered by
5
Answer:
i think answer should be 6!
Step-by-step explanation:
because 5 orators are being taken everytime in random manner. but "ABC" is behaving like a single word according to given condition. so here orators are in order like "{ABC}DEFGH" here total 6 elements , so answer will be 6!
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