Math Question

[Ferdowsi]

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I saw this problem and couldn't figure it out
so here is the ques:
from a group of 3 violinists and 4 pianists, a judge must select 2 violinists and 2 pianists to perform at a music recital. How many different combinations of musicians might perform at recital?


and there is another one close to that:
from a group of 3 singers and 3 comedians, a show organizer must select 2 singers and 2 comedian to appear one after another in a show. how many different ways can the organizer arrange performers for the show?

it's so frustrating 🙁

 

[vvvv]

I saw this problem and couldn't figure it out
so here is the ques:
from a group of 3 violinists and 4 pianists, a judge must select 2 violinists and 2 pianists to perform at a music recital. How many different combinations of musicians might perform at recital?


and there is another one close to that:
from a group of 3 singers and 3 comedians, a show organizer must select 2 singers and 2 comedian to appear one after another in a show. how many different ways can the organizer arrange performers for the show?

it's so frustrating 🙁

first one. 2C3 * 2C4 =
second: 2C3 * 2C3 =

 

[Ferdowsi]

first one. 2C3 * 2C4 =
second: 2C3 * 2C3 =

I don't get it..wat's C

but the answer for the first one is 18
and the second one is 216

 

[vvvv]

I don't get it..wat's C

but the answer for the first one is 18
and the second one is 216

C is combination. because the order in this case is not matter so we use C. formular of C is nCm = m!/n! (m-n)1.
so 2C3 * 2C4 = 3!/2!*1! * 4!/2! * 2! = 18

the second one I dont think you give the right answer.

 

[Ferdowsi]

C is combination. because the order in this case is not matter so we use C. formular of C is nCm = m!/n! (m-n)1.
so 2C3 * 2C4 = 3!/2!*1! * 4!/2! * 2! = 18

the second one I dont think you give the right answer.

thanks for the formula.but for the second one it is the rite answer...i don't know how

 

[vvvv]

thanks for the formula.but for the second one it is the rite answer...i don't know how

If you think about the second one, it should be lower number than the first one because 4 compare to 3. if you could count them out, you will see the answer is not sounded right

 

[Streetwolf]

I saw this problem and couldn't figure it out
so here is the ques:
from a group of 3 violinists and 4 pianists, a judge must select 2 violinists and 2 pianists to perform at a music recital. How many different combinations of musicians might perform at recital?


and there is another one close to that:
from a group of 3 singers and 3 comedians, a show organizer must select 2 singers and 2 comedian to appear one after another in a show. how many different ways can the organizer arrange performers for the show?

it's so frustrating 🙁

First off vvvv you do the big number C the little number, so you do (3 C 2) for example (read "three choose two"). That's just been buggin' me a little bit. 😛

First one is asking for the musicians that will perform so it's a combination. You do (3 C 2) for the violinists and (4 C 2) for the pianists. This picks the 2 violinists and the 2 pianists. You get (3 C 2) = 3 and (4 C 2) = 6 so 3*6 = 18.

Second one is asking for the total number of arrangements. Order MATTERS because the organizer wants the number of ways to arrange a lineup for the show. This is technically a permutation but the easiest way to do it is to use a combination.

You have three comedians and three singers. You want two of each. First determine the number of ways you can select who performs. You do (3 C 2) and (3 C 2) = 3*3 = 9. So you have 9 different groups of 4 people that you could select, without regard to order. Now you want to place them in order. For any group of 4 people, you can order them in 4! = 24 ways. So you take 9*24 = 216 and that's your answer.

 

[vvvv]

First off vvvv you do the big number C the little number, so you do (3 C 2) for example (read "three choose two"). That's just been buggin' me a little bit. 😛

First one is asking for the musicians that will perform so it's a combination. You do (3 C 2) for the violinists and (4 C 2) for the pianists. This picks the 2 violinists and the 2 pianists. You get (3 C 2) = 3 and (4 C 2) = 6 so 3*6 = 18.

Second one is asking for the total number of arrangements. Order MATTERS because the organizer wants the number of ways to arrange a lineup for the show. This is technically a permutation but the easiest way to do it is to use a combination.

You have three comedians and three singers. You want two of each. First determine the number of ways you can select who performs. You do (3 C 2) and (3 C 2) = 3*3 = 9. So you have 9 different groups of 4 people that you could select, without regard to order. Now you want to place them in order. For any group of 4 people, you can order them in 4! = 24 ways. So you take 9*24 = 216 and that's your answer.

excellent.