Question 1058727
.
In how many ways can 2 men, 4 women, 3 boys and 3 girls be selected from 6 men, 8 women, 4 boys and 5 girls if no restrictions are imposed
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<pre>
{{{C[6]^2}}} = {{{(6*5)/(1*2)}}} = 3*5 = 15 ways for men.


{{{C[8]^4}}} = {{{(8*7*6*5)/(1*2*3*4)}}} = 7*2*5 = 70 ways for women.


{{{C[4]^3}}} = 4 ways for boys.


{{{C[5]^3}}} = {{{(5*4*3)/(1*2*3)}}} = 5*2 = 10 ways for girls.


In total, 15*70*4*10 ways.
</pre>

This problem is about <U>combinations</U>. 


On combinations, see the 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>

in this site.


Also, 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 "<U>Combinatorics: Combinations and permutations</U>".