5 girls and 3 boys are arranged randomly in a row. Find the probability that there is one boy on each end.
A PS Question on Permutations and Combinations
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6/56 = 3/28
3p2 * 6! = 720*6 / 8! = 3/28
DiD,
I liked your approach. The answer is indeed 3/28.
I am still mentioning another “equivalent” approach..
The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3
the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2
the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways
Total possible outcome = 8!
Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28
2*(3C2 * 6!/8!) = 3/28
Three boys can be arranged at each ends in 3*2 ways = 6 ways
Now there are 6 position & 5 girls, 1 boy can be arranged in 6! ways.
Total no of arrangement from 3 Boys, 5 girls = 8!
Probability = 6*6!/8! = 6/58 = 3/28
3/28
1st point: There are 8! possible orders for the 5 girls and 3 boys;
2nd point: There are 36, or 6^2, ways for the boys to be arranged:
a. There are 3 ways for 1 boy to not be on the end while 2 boys are on the ends.
b. The 2 boys on the ends, can be switch/exchange places. So, for every position the 1 boy not on the end is in, there are 2 possibilities for the boys at the ends.
c. The 1 boy not on the end can be in any 1 of 6 non-end positions.
d. Conclusion: (1 of 3 not on the ends) x (2 boys in 2 positions on the ends) x (6 positions for the 1 boy not on an end) = 3 x 2 x 6 = 36
3rd point: There are 5! arrangements for the 5 girls for each of the 36 arrangements of the boys.
Conclusion: (6^2 x 5!)/8!=(6 x 6)/(8 x 7 x 6)= 6/(8 x 7)=6/56=3/28
2! * 6! / 8 !
(3c2 X 2 X 6! ) divided by 8! = 3/28
two boys to sit at the ends can be selected in 3c2 ways and then can made to sit at the either ends so we have X 2 ; remaining 6 can be arranged in 6! ways so we have (3c2 X 2 X 6! ) as possible favourable outcomes against 8! , total number of possible permutations.