Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated ?
Question:
Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated ?
Correct Option: , 2
Solution:
Total numbers in three familes $=3+3+4=10$ so total arrangement $=10$ !
Favourable cases
$=$ $3 !$ $3 ! \times 3 ! \times 4 !$
$\therefore$ Probability of same family memebers are
together $=\frac{3 ! 3 ! 3 ! 4 !}{10 !}=\frac{1}{700}$
so option(2) is correct.
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