Question 1181974: Ron is having a dinner party. He invites 7 out of his 10 friends. In how many ways can he choose his guests if two of them, Al and Sue, must attend together or not at all?
Answer by ikleyn(52807)   (Show Source):
You can put this solution on YOUR website! .
Ron is having a dinner party. He invites 7 out of his 10 friends.
In how many ways can he choose his guests if two of them, Al and Sue, must attend together or not at all?
~~~~~~~~~~~
As you understand, the answer is the sum of two quantities.
One quantity is the number of the ways, when both Al and Sue attend the party.
This quantity is equal to  =  =  = 8*7 = 56
(because you simply select 5 guests from the 10-2 = 8 friends to add them to Al and Sue).
The other quantity is the number of the ways, that exculde both Al and Sue from the attendees.
This quantity is  =  = 8.
(because this time you select 7 friends not from 10 persons, but from 10-2 = 8 persons, only).
The final ANSWER is 56 + 8 = 64 ways.
Solved, answered and carefully explained.
----------------
This problem is on COMBINATIONS.
On Combinations, see the lessons
- Introduction to Combinations
- PROOF of the formula on the number of Combinations
- Problems on Combinations
- Problems on Combinations with restrictions
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
|
|
|
