document.write( "Question 1209003: Suppose we want to choose 6 letters, without replacement, from 8 distinct letters.
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document.write( " How many ways can this be done, if the order of the choices does not matter?\r
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document.write( "How many ways can this be done, if the order of the choices matters? \n" );
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Algebra.Com's Answer #847569 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "This might be a bit strange, but I'll address the \"order matters\" portion first.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are n = 8 letters and r = 6 slots to fill. \n" ); document.write( "We have 8 choices for the first slot, then 7 for the next, and so on until reaching n-r+1 = 8-6+1 = 3 choices for the 6th slot.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Multiplying out these values gives: (8*7*6)*(5*4*3) = 20160 different permutations where order matters.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Another way to reach this value is to use the nPr permutation formula. \n" ); document.write( "nPr = (n!)/( (n-r)! ) \n" ); document.write( "which I'll let the student handle the scratch work if s/he chooses this route. \n" ); document.write( "The exclamation mark indicates a factorial.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-----------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now onto the \"order doesn't matter\" portion. \n" ); document.write( "For any group of 6 items, we have 6! = 6*5*4*3*2*1 = 720 ways to rearrange said group. \n" ); document.write( "This means that the previous value we found (20160) is too large by a factor of 720 if order doesn't matter. \n" ); document.write( "We'll divide the permutation value over r! = 6! = 720 to find the combination value.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A very useful formula is \n" ); document.write( "nCr = (nPr)/(r!) \n" ); document.write( "which is a way to connect permutations to combinations. \r \n" ); document.write( "\n" ); document.write( "There are 20160/720 = 28 combinations.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This value can be found by looking at Pascal's Triangle. \n" ); document.write( "Since n = 8, we start with the row \"1,8,...\" \n" ); document.write( "r = 0 is the starting index and refers to the left-most value in this row. Increment r by 1 each time you move to the right. Once reaching r = 6, you should arrive at 28 in Pascal's Triangle.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Or if you wanted you can use this nCr formula \n" ); document.write( "nCr = (n!)/( r!*(n-r)! ) \n" ); document.write( "I'll let the student handle the scratch work if s/he chooses this route. \n" ); document.write( "Notice the presence of n! up top and (n-r)! down below; both of which are found in the nPr formula. \n" ); document.write( "So that can help explain why nCr = (nPr)/(r!) is true.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-------------------------------------------------------------------------- \n" ); document.write( "--------------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answers:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "How many ways can this be done, if... \n" ); document.write( "Order doesn't matter? 28 \n" ); document.write( "Order does matter? 20160 \n" ); document.write( " \n" ); document.write( " |