<p>One thing I dont understand in Combination is that how can you make the Combination equation out of the Permutation equation?
nCr= nPr!/ r!</p>
<p>OK, let’s start with a permutation question: Say you want to make 3-letter codes out of the first 10 letters, no repeats…that is a classic permutation question and the answer is 10P3 or 10x9x8=720</p>
<p>Now suppose the question is changed so that you want to make 3-letter groupings. The answer is not still 720. Your previous answer has counted different orders as different codes, but they would not make different groupings. So for example, </p>
<p>ABC ACB BAC BCA CAB CBA are all different codes but the same group of letters. So the answer 720 has overcounted the groupings…by a factor of 6 (in this case). But where did the 6 come from? It is the number of ways to arrange the 3 letter group. So 3! tells you the factor by which you have overcounted. Thus, 10C3 = 10P3/3!</p>
<p>Hope that helps…</p>
<p>nPr = n!/(n-r)!
nCr = n!/[(n-r)! r!] which is the same thing as nPr/r!</p>