Respuesta :
Answer:
(a) 65318400
(b) 1080
(c) 311040
Step-by-step explanation:
I think you forgot to put the questions
(a) if the programmes can be perfomed in any order ?
(b) if the programmes of the same kind were perfomed at a stretch?
(c) if the programmes of the same kind were perfomed at a strech and considering the order of performance of the programmes of the same kind?
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The number of choices available to them were [tex]65318400[/tex] , [tex]1080[/tex] and [tex]311040[/tex].
What is combination ?
Combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter.
Combination [tex](^nC_{r} ) = \frac{n ! }{r!(n-r)!}[/tex]
1.
We have, to perform [tex]2[/tex] short plays, [tex]4[/tex] recitals and [tex]3[/tex] dance programmes in any order.
So,
The number of ways to select [tex]2[/tex] short plays out of [tex]3[/tex] using the mentioned formula,
[tex](^3C_{2} ) = \frac{3 ! }{2!(3-2)!}[/tex]
[tex]= \frac{3 *2*1 }{2*1*1}=3[/tex]
Now,
The number of ways to select [tex]4[/tex] recitals out of [tex]6[/tex] using the mentioned formula,
[tex](^6C_{4} ) = \frac{6 ! }{4!(6-4)!}[/tex]
[tex]= \frac{6*5*4 ! }{4!*2*1} =15[/tex]
Now,
The number of ways to select [tex]3[/tex] dance programs out of [tex]4[/tex] using the mentioned formula,
[tex](^4C_{3} ) = \frac{4 ! }{3!(4-3)!}[/tex]
[tex]= \frac{4*3 ! }{3!*1!}=4[/tex]
Now,
The number of ways to arrange [tex]9[/tex] program using the mentioned formula,
[tex](^9P_{9} ) = \frac{9 ! }{(9-9)!}[/tex]
[tex]=9![/tex]
[tex]=9*8*7*6*5*4*3*2*1=362880[/tex]
So, the total number of ways by multiplying the values
i.e. [tex]362880*4*15*3=65318400[/tex]
So, the total number of ways to perform the programs are [tex]65318400[/tex] .
2.
We have, to perform the programmes of the same kind.
So,
The number of ways to arrange [tex]3[/tex] different program using the mentioned formula,
[tex](^3P_{3} ) = \frac{3 ! }{(3-3)!}=3!=3*2*1=6[/tex]
So, the total number of ways by multiplying the values,
i.e. [tex]6*4*15*3=1080[/tex]
3.
We have to perform, the programmes of the same kind and considering the order of performance of the programmes of the same kind.
So,
The number of ways to select [tex]2[/tex] short plays out of [tex]3[/tex] using the mentioned formula,
i.e. [tex](^3P_{2} ) = \frac{3 ! }{(3-2)!}=\frac{3!}{1!} =3*2=6[/tex]
Now,
The number of ways to select [tex]4[/tex] recitals out of [tex]6[/tex] using the mentioned formula,
[tex](^6CP_{4} ) = \frac{6 ! }{(6-4)!}=\frac{6!}{2!} =\frac{6*5*4*3*2!}{2!} =360[/tex]
Now,
The number of ways to select [tex]3[/tex] dance programs out of [tex]4[/tex] using the mentioned formula,
[tex](^4P_{3} ) = \frac{4 ! }{(4-3)!}=\frac{4!}{1!}=4*3*2=24[/tex]
So,
The number of ways to arrange [tex]3[/tex] different program using the mentioned formula,
[tex](^3P_{3} ) = \frac{3 ! }{(3-3)!}=3!=3*2*1=6[/tex]
So, the total number of ways by multiplying the values,
i.e. [tex]6*24*360*6=311040[/tex]
Hence, we can say that the number of choices available to them were [tex]65318400[/tex] , [tex]1080[/tex] and [tex]311040[/tex].
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