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

Question

Dr. Abubacker · Accepted Answer

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

Q. 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

Explanation

Reviewed by Dr. Abubacker

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