• # question_answer Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is A) $\frac{10\ !}{6}$ B) $3\ !\ 7\ !$ C) $^{10}{{P}_{3}}\ .\ 7\ !$ D) None of these

Correct Answer: A

Solution :

For A, B, C to speak in order of alphabets, 3 places out of 10 may be chosen first in $1\ .{{\ }^{3}}{{C}_{2}}=3$ways. The remaining 7 persons can speak in $7\ !$ ways. Hence, the number of ways in which all the 10 person can speak is $^{10}{{C}_{3}}\ .\ 7\ !\ =\frac{10\ !}{3\ !}.=\frac{10\ !}{6}.$

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