You visited us 1 times! Enjoying our articles? Unlock Full Access!

Permutation

15 persons wh...

Question

15 persons which include A, B, C are to speak at a function. Find the number of ways in which it can be done, if A wants to speak before B and B wants to speak before C.

^{12} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{15} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{15} C_{3} \times 12!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^{15} C_{3} \times 3!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $^{15} C_{3} \times 12!$
Here, places for A, B and C can be chosen in $^{15} C_{3}$ ways. Now, 12 person remain to speak and these 12 persons can speak in 12! ways. Hence, number of way in which they can speak is $^{15} C_{3} \times 12!$ .

Suggest Corrections

Similar questions

Q. 15 persons which include A, B and C are to speak at a function. Find the number of ways in which it can be done, if A wants to speak before B and B wants to speak before C.

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

Q. Ten persons, amongst whom are

A, B

and

C

are 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

A, B, C

are three persons among

7

persons who speak at a function. The number of ways in which it can be done. If '

A

' speaks before '

B

' and '

B

' speaks before '

C

' is

A, B, C

are three persons among

7

persons who speak at a function. The number of ways in which it can be done if

^{'} A^{'}

speaks before

^{'} B^{'}

and

^{'} B^{'}

speaks before

^{'} C^{'}

Join BYJU'S Learning Program

15 persons which include A, B, C are to speak at a function. Find the number of ways in which it can be done, if A wants to speak before B and B wants to speak before C.

The correct option is C 15C3×12!Here, places for A, B and C can be chosen in 15C3 ways. Now, 12 person remain to speak and these 12 persons can speak in 12! ways. Hence, number of way in which they can speak is 15C3×12!.

The correct option is C $^{15} C_{3} \times 12!$
Here, places for A, B and C can be chosen in $^{15} C_{3}$ ways. Now, 12 person remain to speak and these 12 persons can speak in 12! ways. Hence, number of way in which they can speak is $^{15} C_{3} \times 12!$ .