3.IN HOW MANY WAYS 3 BOYS AND 5 GIRLS BE ARRANGED IN A ROW SO THAT i)NO TWO BOYS R TOGETHER ii)ALL THE GILS ARE TOGETHER.(WITH EXPLANATION )
Asked by Satyam sinha | 13th Feb, 2014, 09:06: PM
i) 5 girls can be seated in a row in 5P5 = 5! ways.
Now, in the 6 gaps, 3 boys can be arranged in 6P3 ways.
Hence, the number of ways in which no two boys sit together
= 5! x 6P3 = 120 x 120 = 14400 ways
ii) Assume 5 girls to be together i.e. (one). Now there are 4 students.
Possible ways of arranging them are 4! ways
Now 5 girls can be arranged among themselves in 5! ways.
Thus, total no. of ways in which all girls are together = 4! x 5! = 24 x 120 = 2880
Answered by | 14th Feb, 2014, 12:24: PM
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