document.write( "Question 883310: Please help me on this. This is confusing (it's solution) $\"%28a%2Bb%2Bc%2Bd%2Be%29%5E3\"$ Thanks In Advance \n" ); document.write( "

Algebra.Com's Answer #533457 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
DISCLAIMERS:
\n" ); document.write( "I am not sure what the question is (what is expected).
\n" ); document.write( "Maybe there is a more elegant solution.
\n" ); document.write( "
\n" ); document.write( "MY BEST TRY:
\n" ); document.write( "To calculate $\"%28a%2Bb%2Bc%2Bd%2Be%29%5E2\"$ or $\"%28a%2Bb%2Bc%2Bd%2Be%29%5E3\"$ you can clench your teeth and start applying the distributive property,
\n" ); document.write( "or you can think of what multiplication means and try to figure out what the terms of the product should be.
\n" ); document.write( " $\"%28A%2BB%2BC%2BD%2BE%29%28a%2Bb%2Bc%2Bd%2Be%29\"$
\n" ); document.write( "The product will have terms that represent all the possible products you could get multiplying one term of $\"%28A%2BB%2BC%2BD%2BE%29\"$ times one term of $\"%28a%2Bb%2Bc%2Bd%2Be%29\"$ .
\n" ); document.write( "There are many possibilities.
\n" ); document.write( "It is a question of combinations.
\n" ); document.write( "It is like choosing one drink from a 5-item list, and one food from a 5-item menu.
\n" ); document.write( "It is like choosing one of 5 available shorts, and one of 5 available T-shirts.
\n" ); document.write( "For $\"%28A%2BB%2BC%2BD%2BE%29%28a%2Bb%2Bc%2Bd%2Be%29\"$ , you get $\"25=5%2A5\"$ different products.
\n" ); document.write( "Five of the products are same letter products:
\n" ); document.write( " $\"Aa\"$ , $\"Bb\"$ , $\"Cc\"$ , $\"Dd\"$ and $\"Ee\"$ .
\n" ); document.write( "The other 20 products are possible combinations of 2 different letters.
\n" ); document.write( " $\"Ab\"$ and $\"aB\"$ are 2 of them.
\n" ); document.write( "(I like to write the factors in alphabetical order, but $\"aB=Ba\"$ ).
\n" ); document.write( "There are $\"5%2A4=20\"$ ways of making those combinations of 2 different letters,
\n" ); document.write( "pairing each of the 5 capital letters with one of the 4 different lowercase letters.
\n" ); document.write( "When the two factors are the same,
\n" ); document.write( "as in $\"%28a%2Bb%2Bc%2Bd%2Be%29%5E2=%28a%2Bb%2Bc%2Bd%2Be%29%28a%2Bb%2Bc%2Bd%2Be%29\"$ ,
\n" ); document.write( "the 20 products with 2 different letters come in matching pairs,
\n" ); document.write( "because $\"Ab\"$ and $\"aB\"$ both turn into $\"ab\"$ .
\n" ); document.write( "You get $\"ab\"$ twice, and the same goes for all possible combinations of 2 different letters.
\n" ); document.write( "

\n" ); document.write( "Writing $\"2%28ab%2Bac%2Bad%2Bae%2Bbc%2Bbd%2Bbe%2Bcd%2Bce%2Bde%29\"$ is more efficient than writing each 2-letter product twice.
\n" ); document.write( "
\n" ); document.write( "For

,
\n" ); document.write( "with 5 terms in each of the 3 factors you would get
\n" ); document.write( " $\"5%2A5%2A5=5%5E2=125\"$ products, but there are many repeats.
\n" ); document.write( "There is just one way to get $\"a%5E3\"$ , $\"b%5E3\"$ , $\"c%5E3\"$ , $\"d%5E3\"$ and $\"e%5E3\"$ ,
\n" ); document.write( "but there are several ways to make $\"abc\"$ ,
\n" ); document.write( "and there are several ways to make many other products like $\"a%5E2b\"$ and $\"ab%5E2\"$ .
\n" ); document.write( "Rather than writing all repeats of the same product, we can list them grouped, sort of like $\"%28a%2Bb%2Bc%2Bd%2Be%29%5E2\"$ above.
\n" ); document.write( "
\n" ); document.write( "There are $\"10\"$ possible 3-different-letter combination products, and each one will appear repeated $\"6\"$ times,
\n" ); document.write( "accounting for $\"6%2A10=60\"$ of the $\"125\"$ products.
\n" ); document.write( "How do we know there are $\"10\"$ possible 3-different-letter combinations?
\n" ); document.write( "Your teacher may say that it is combinations of 5 taken 3 at a time,
\n" ); document.write( "and that you can calculate that as $\"5%21%2F%285-3%29%213%21=5%2A4%2A3%2F%282%2A3%29=10\"$ ,
\n" ); document.write( "but it is easy enough to list them and count them.
\n" ); document.write( "Where does the $\"6\"$ times come from?
\n" ); document.write( "How many ways can we make $\"abc\"$ ?
\n" ); document.write( "You could choose the $\"a\"$ from the first $\"%28a%2Bb%2Bc%2Bd%2Be%29\"$ factor, or from the second one, or from the third one.
\n" ); document.write( "After that you would still have two choices for the $\"b\"$ ,
\n" ); document.write( "and there would be $\"3%2A2=6\"$ ways to make $\"abc\"$ .
\n" ); document.write( "Your teacher may say that it is permutations of 3, or $\"3%21\"$ .
\n" ); document.write( "The same goes for $\"abd\"$ , $\"abe\"$ , $\"acd\"$ , etc.
\n" ); document.write( "
\n" ); document.write( "As for products like $\"a%5E2b\"$ , there are $\"10\"$ possible sets of 2 letters, but the same 2 letters can make 2 such products, so there are $\"10%2A2=20\"$ such products.
\n" ); document.write( "Each of those products can be made $\"3\"$ different ways,
\n" ); document.write( "because there are 3 factors where you can find the letter that is not squared.
\n" ); document.write( "That accounts for $\"3%2A20=60\"$ of the $\"125\"$ products.\r
\n" ); document.write( "\n" ); document.write( "All in all, counting all repeats, we have
\n" ); document.write( " $\"5\"$ cubes,
\n" ); document.write( " $\"60\"$ factors with one letter squared, and
\n" ); document.write( " $\"60\"$ factors with no exponents,
\n" ); document.write( "adding up to the $\"125\"$ expected factors.
\n" ); document.write( "

\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );