The number of ways of dividing 10 girls into two groups of 5 each so that two shortest girls are in the different groups is:
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answer is 1.
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Detailed Solution
Excluding two shortest girls, the remaining eight girls can be divided into two groups in 12! 8C4 4C4=35 ways. The two shortest girls can be put in two different groups in 2 ways. Therefore, the required number of ways is 70.