Question 1183476
.
A student has 12 classmates.
(a) In how many combinations can she invite five of them to lunch?
(b) Two classmates are having a dispute and refuse to be together. 
In how many combinations can she invite five classmates if these two are not together?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Part &nbsp;(a)</U>



<pre>
    There are  {{{C[12]^5}}} = {{{12!/(5!*(12-5)!)}}} = {{{12!/(5!*7!)}}} = {{{(12*11*10*9*8)/(1*2*3*4*5)}}} = 792 such combinations.    <U>ANSWER</U>
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Part &nbsp;(b)</U>



<pre>
    {{{C[10]^5}}}  combinations of five, where NEITHER classmate A NOR classmate B present (the selection is 5 from 10),  P L U S


    {{{C[10]^4}}}  combinations of four (+ A), where classmate A does present, but classmate B does not (the selection is 4 from 10),  P L U S


    {{{C[10]^4}}}  combinations of four (+ B), where classmate A does not present, but classmate B does (the selection is 4 from 10).


In all, there are  {{{C[10]^5}}} + {{{2*C[10]^4}}} = 252 + 2*210 = 672 such combinations.    <U>ANSWER</U>



It can be computed by different way as  {{{C[12]^5}}} - {{{C[10]^3}}} = 792 - 120 = 672  (giving the same answer),


taking all possible combinations of 5 from 12 and subtracting all combinations of 5 from 12, that include both disputed classmates.



    +-----------------------------------------------------------------------+
    |   CONSIDER this parallel computing  AS a GOOD and a NECESSARY CHECK   |
    +-----------------------------------------------------------------------+
</pre>


Solved.



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On Combinations, &nbsp;see introductory lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-combinations.lesson>PROOF of the formula on the number of Combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Combinations.lesson>Problems on Combinations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Combinations-problems-with-restrictions.lesson>Problems on Combinations with restrictions</A> 

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic &nbsp;"<U>Combinatorics: Combinations and permutations</U>". 



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.