Hello raja987654321

 

Please post only one querry at once.

 

(i) As per your query there are seven blank spaces and B in the first position

then the query loks like as :

 

1	2	3	4	5	6	7
B

A

B

C

A

B

C

 

In this arrangement the letters in first batch as well as in second batch can be arranged as 3! manner so total arrangements are 3! * 3! = 36 arrangements.

 

Now as in the first set there are two cases where C will be at last place and number of arragements where C be at starting position in second sets are 2! = 2 so there are 2*2 = 4 possible arrangements where C must be come together. Same way when B at last place and A at last place in first set then there are again 2 arrangements possible where each letter at startig in the second set so totol 12 arrangements are possible where letters occupy consecutive positions.

 

Now total requisite arrangements are 36 - 12 = 24

 

Again in the first set there are 2! ways where B in the Starting position so we should avoid these cases too.