Question 364413: how many methods are there to sit 4 boys and 4 girls alternatively in a row if a one boy and a one girl should sit always together?
Found 2 solutions by sudhanshu_kmr, chandrumail:
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
let assume that particular boy and girl that should sit together only one entity.
3 boys can sit in 3 place by 3! ways
3 girls can sit in 3 places by 3! ways..
no. of ways by which pair can sit in a row of 8 people = 7 *2 = 14
total no. of ways = 14*6*6 =504
answer is 504
Answer by chandrumail(4)  (Show Source):
You can put this solution on YOUR website! Let's assume the 4 girls are G1, G2, G3, G4 and the 4 boys B1, B2, B3, B4.
Let G1,B1 be the pair that needs to sit always together
As per problem, the boys and girls should sit alternatively.
The possibile scenarios are:
G1B1 G2B2G3B3G4B4 (girl, boy, girl, boy....)
B1G1 B2G2B3G2B4G4 (boy, girl, boy, girl....)
In the 1st arrangement, G2,G3 and G4 can be arranged in 3! ways and B2,B3 and B4 can be arranged in 3! ways
Hence, total no of ways the 3 girls and 3 boys can be arranged is 3! * 3! = 36 ways
The G1B1 doublet can be placed in any of the 7 positions (4 positions as G1B1 and 3 positions as B1G1 satisfying the condition that boys and girls should sit alternatively)
So, the number of arrangements = 7*36 = 252
We have another set of 252 arrangements for the 2nd scenario
B1G1 B2G2B3G2B4G4 (boy, girl, boy, girl....)
So, total no of ways = 252 + 252 = 504 ways
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