Tricky math!

Let the companies be A,B,C,D,E and A and B are the Bombay companies. Now it is fixed that 2 persons wants to go to Bombay companies and three persons wants to go to C or D or E. From the remaining 5 people we choose 2 persons to be sent to Bombay companies i.e A or B and the rest 3 are sent to C or D or E. This can be done in 5C2 ways. Now the 4 people to be sent to Bombay and 6 people to be sent outside is fixed. Now observe the 4 persons who are to be sent to Bombay. They can be arranged in 4! ways, and let the first 2 person in each arrangement be sent to company A and next 2 in company B.[for example- For the arrangement P2,P3,P1,P4 A-(P2,P3) B-(P1,P4)].Now there are repetations like P2,P3,P1,P4 and P3,P2,P1,P4 may seem to be a different arrangement but after grouping they are same i.e in the first case A-(P2,P3) B-(P1,P4) and in the second case A-(P3,P2) B-(P1,P4).hence both the cases are same. So this repetation takes place for company A in 2! ways and for company B in 2! ways. So the number of ways are 4!/(2!)^2. Same as the outside companies-6!/(2!)^3. did you understand or will I make it more clear? AND BY THE WAY DON'T CALL ME SIR I AM YOUNGER TO YOU. I AM JUST A 15 YEARS OLD BOY!!!!!!!!