Question 1181512
.
How many "words" can be made from letters of the word CATERING using different permutations, 
according to the following rules.
1. Two vowels and two consonants must be used without repetition.
2. Three vowels and one consonant must be used, without repetition.
3. At least two vowels and at least two consonants must be used, without repetition, in six-letter words.
The answer for 1 is too 720. Answer for 2 is 120 and answer for 3 is 18 000. 
I know the answers but I don't know how to work it out.
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<pre>
First notice that there are 8 letters in all; of them, 3 are vowels and 5 are consonants.


(1)  We can select 2 vowels     from 3 vowels     by  {{{C[3]^2}}} = {{{(3*2)/2}}} =  3 different ways.
 
     We can select 2 consonants from 5 consonants by  {{{C[5]^2}}} = {{{(5*4)/2}}} = 10 different ways.

     We can combine these sets in 2*10 = 20 different ways.

     Finally, we can order (arrange) each of these sets in 4! = 4*3*2*1 = 24 different ways.


     Applying the fundamental counting principle, we get the total number of possible combinations as 

                  3*10*24 = 30*24 = 720.   <U>ANSWER</U>




(2)  We can select 3 vowels     from 3 vowels     by  {{{C[3]^3}}} = 1 different ways.
 
     We can select 1 consonants from 5 consonants by  {{{C[5]^1}}} = 5 different ways.

     We can combine these sets in 1*5 = 5 different ways.

     Finally, we can order (arrange) each of these sets in 4! = 4*3*2*1 = 24 different ways.


     Applying the fundamental counting principle, we get the total number of possible combinations as 

                  1*5*24 = 5*24 = 120.   <U>ANSWER</U>




(3)  In this case, we combine 2 vowels and 4 consonants  OR  3 vowels and 3 consonants - and we have no other opportunities.


     Applying the same logic, we have the formula

        total number of arrangements = {{{6!*(C[3]^2*C[5]^4 + C[3]^3*C[5]^3)}}} = 720*(3*5 + 1*10) = 720*25 = 18000.    <U>ANSWER</U>
</pre>

Solved.


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This problem is on COMBINATIONS.


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/Fundamental-counting-principle-problems.lesson>Fundamental counting principle problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/OVERVIEW-the-lessons-on-Permutations-and-Combinations.lesson>OVERVIEW of lessons on Permutations and Combinations</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.



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