SATHEE: Permutations And Combinations Question 29

Permutations And Combinations Question 29

Question: 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

Options:

A) $ \frac{10\ !}{6} $

B) $ 3\ !\ 7\ ! $

C) $ ^{10}P_3\ .\ 7\ ! $

D) None of these

Show Answer

Answer:

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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Permutations And Combinations Question 29

Question: 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

Options:

Answer:

Solution:

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