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}

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^{15} C_{3}

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

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

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

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