# 2 question in permutations and Combination

• Feb 25th 2010, 04:03 AM
r-soy
2 question in permutations and Combination
1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
A ) how many games must be scheduled ?
B ) How many games must be on each team ?

A ) 10C9 = 10
B ) 10C1 = 10

2 )_A catering service offers 8 appetize, 10 main courses and 7 desserts . A Chairperson is to select
3 appertizs , 4 main courses and 2 desserts for A dinner . How many way can this be done ?

Here in this question we use C or P and why ?
• Feb 25th 2010, 04:18 AM
Quote:

Originally Posted by r-soy
1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
A ) how many games must be scheduled ?
B ) How many games must be on each team ?

A ) 10C9 = 10
B ) 10C1 = 10

2 )_A catering service offers 8 appetize, 10 main courses and 7 desserts . A Chairperson is to select
3 appertizs , 4 main courses and 2 desserts for A dinner . How many way can this be done ?

Here in this question we use C or P and why ?

hi

(2) That would be combination because take dessert for example , it can be ice cream , pudding , cakes , ... the order is not important here , it could be ice cream , cakes , pudding , OR pudding , ice , cakes , its the same dessert we are talking about .

so 8C3 x 10C4 x 7C2
• Feb 25th 2010, 04:49 AM
Quote:

Originally Posted by r-soy
1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .
A ) how many games must be scheduled ?
B ) How many games must be on each team ?

A ) 10C9 = 10
B ) 10C1 = 10

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(A)

TeamA plays the other 9 teams

TeamB has played TeamA and must play the other 8 teams

TeamC has played both TeamA and TeamB, so has the other 7 teams to play.

This is 9+8+7+6+5+4+3+2+1=45 games.

Or...each time, 5 games can be played (say on a Saturday).
Each team must play 9 games.

That's 9(5)=45 games
• Feb 26th 2010, 09:50 AM
r-soy
But in like this queation we solve by rule

i mean we use C
for example 10 C 9

i want expline the solution like this way
• Feb 26th 2010, 10:30 AM
Quote:

Originally Posted by r-soy
But in like this queation we solve by rule

i mean we use C
for example 10 C 9

i want expline the solution like this way

Ok,

in that case, in terms of "selections",
we need to calculate the number of ways of pairing off 2 teams from 10.

That will give us the number of games playable in a league of 10 teams.

$\binom{10}{2}=10c_2=\frac{10!}{(10-2)!2!}=\frac{10!}{8!2!}=\frac{10(9)}{2}=5(9)=45$
• Feb 27th 2010, 03:15 AM
Quote:

Originally Posted by r-soy
1 ) there are 10 teams in a league .if each team is to play evcry other team exactly once .

B ) How many games must be on each team ?

B ) 10C1 = 10

Hi r-soy,

could you be more specific when you say "on each team" ?

If so, this is $9c_9,\ or\ \binom{9}{9}$ as each team plays the 9 others (they cannot play against themselves).