web2.0calc
 
+0  
 

Odds

-2
850
4
avatar+343 

I have 2 questions im new at odds.

A girl have 8 friends and she want to invite 5 of them to her house.How many choices have if 2 of her friends are quarreled and can't invite both of them in her house?

 

And the other:

We have 10 people and we want put them in 10 seatmate chairs but 2 of them must have difference 3 chairs.How much choices we have to put them?

Thank you.

 Oct 13, 2019

4+0 Answers

 #1
avatar+6248 
+1

....

.
 Oct 13, 2019
edited by Rom  Oct 18, 2019
 #3
avatar+343 
0

Thank you! Rom

Dimitristhym  Oct 13, 2019
 #2
avatar+447 
+4

#1There are (63)=20(63)=20 possible ways for inviting the two. Also, there are (65)=6(65)=6 ways of not inviting them. Thus, there are 26 possible invitations that satisfy these criteria ((63)+(65)=26(63)+(65)=26) \(26\)

 

#2 use combinations for this one also

 Oct 13, 2019
 #4
avatar+343 
-2

2nd Find it \(84*8!\)

 Oct 13, 2019

0 Online Users

Top Users

avatar+128408 CPhill moderator
avatar+118608 Melody moderator
avatar+33615 Alan moderator
avatar+14905 asinus moderator
avatar+9519 MaxWong
avatar+3144 admin administrator
avatar+2440 GingerAle
avatar+1763 tomtom
avatar+1484 bader
avatar+1037 parmen
avatar+1030 blackpanther

Sticky Topics