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

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) (10 !/6) (B) 3 ! 7 ! (C)  10 P 3 ⋅ 7 ! (D) None of these

Tardigrade Admin · Accepted Answer

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 is 10 C 3 ⋅ 7 ! =(10 !/3 !)=(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

$\frac{10 !}{6}$

$3! 7!$

$^{10} P_{3} \cdot 7!$

None of these

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 is $^{10} C_{3} \cdot 7!$
$= \frac{10 !}{3 !} = \frac{10 !}{6}$ .

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

610!​

3!7!

10P3​⋅7!