counting question

- anonymous

counting question

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- schrodinger

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- anonymous

a volley ball league with 4 teams ,each team plays exactly 4 times with each of the other 3 teams in the league .what is the total number of games played in this league? a)4 b)6 c)12 d)16 e)24

- anonymous

@imqwerty

- anonymous

handshake problem times 4

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## More answers

- anonymous

what of meaning of handshake

- anonymous

|dw:1441084213468:dw|

- anonymous

google "handshake problem"

- anonymous

ok thank u

- anonymous

@pgpilot326

- anonymous

do you know about "combinations"??

- anonymous

ok can i solve it by npr or ncr@pgpilot326

- anonymous

combinations are not needed... nor useful in this problem

- anonymous

yes right,it can be very easily solve by combinations

- anonymous

ncr

- anonymous

yes,now try to solve it

- anonymous

how many games are there if each team plays the other teams exactly once?

- anonymous

hassan, you are leading nourage down the wrong path!

- anonymous

ncr is combinations right

- anonymous

@pgpilot326 try to solve it by combinations i think that really works

- anonymous

but first let him solve it,then if he couldn't i will show you

- anonymous

combinations is not the way

- anonymous

ncr is combinations or not

- anonymous

@Nourage and don't forget to multiply the answer with how much these teams gonna play with each other

- anonymous

@Nourage do you know the formula of combinations

- anonymous

please explain why you think combinations is relevant here...

- anonymous

no i don t know hassan

- anonymous

\[nCr=\frac{ n! }{ n!(n-r)! }\]

- anonymous

plz hlp

- anonymous

you know what is n and r?

- anonymous

r radius

- anonymous

or not

- anonymous

did you look up the "handshake problem?" and did you see my drawing?

- anonymous

combinations is ncr in the calc

- anonymous

right or wrong

- anonymous

hassan

- anonymous

nope friend, n is the total number of teams and r is the team which will play just like two people will involve in a handshake two teams will participate to play a match,that's why we take r=2

- anonymous

again, @Hassan3130 explain your reasoning for combinations! it's wrong and you're confusing the issue!

- anonymous

i can t understand

- anonymous

plz help

- anonymous

\[\frac{ 4! }{ 2!2! }=\frac{ 24 }{4 }\]

- anonymous

and wt is c

- anonymous

i am saying to try for combination because i think it's easy to find but if you don't read much about this topic than you can follow what @pgpilot326 said

- anonymous

ok but can u said to me how to solve it by ncr or npr in the calc

- anonymous

@Hassan3130

- anonymous

yes you can solve it from scientific caluclator

- anonymous

how plz

- anonymous

@Hassan3130

- anonymous

but firstly you have to know the value of r and n

- anonymous

r is 2 and n is4

- anonymous

right

- anonymous

yes

- anonymous

then what i can i do on calc

- anonymous

have you find it written on calculator"nCr"

- anonymous

4n2

- anonymous

4c2

- anonymous

6

- anonymous

yes right

- anonymous

then 6*4 right

- anonymous

yes absolutely right

- anonymous

cause they will play with each other 4 times

- anonymous

thank u @Hassan3130 and @pgpilot326

- anonymous

your the best

- anonymous

well done now i hope that you can solve questions like this

- anonymous

sure hassan

- anonymous

look at my 2nd response... see if that makes more sense. combination is valid

- anonymous

thank u @pgpilot326

- anonymous

i m a fan of u @Hassan3130 and pgpilot326

- anonymous

bi hassan

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