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Name: Ch 3 Combinations - 3.1 MDM 4U1 Permutations with Identical Items Notice in the first example, that there are 2! ways or organizing W1 and W2 in each arrangement of WWRG. If the white blocks are treated as identical, then the 21 ways to organize W1 and W2 no longer matter. Therefore, to find the number of distinct (truly different) permutations of WWRG (where the white blocks are no longer treated as being different), we need divide out 2! Suppose you want to know how many ways you can arrange a set of n items. If some number of these items, 'a , are identical, then you need to undo or divide out the al ways you would have arranged those items within the set if they were distinct. If you also had 'b' identical items, you would also divide out the b! ways of organizing those items too. The number of permutations of n objects, of which a are identical items of one kind, b are identical items of a second kind, c are identical items of a third kind, ... and so on is: n! Permutations = a!b!c!... -stop here and check in with your teacher before continuing- C: "NUMBERS" - Distinct objects in a fixed order. How many ways are there to arrange the letters in the word NUMBER if the consonants must remain in their original order? (that is, N, M, B, R) Hint! If the letters NMBR must all be in that order, in each arrangement, then you can pretend that all the consonants, NMBR, are identical items (which we'll call 'z'). In other words, NUMBER becomes zUzzEz . Use this to complete the question! (Why does this work? Because, in each arrangement of 6 letters, there is only 1 way of organizing NMBR that matters, compared to all the ways you can organize the four letters, N,M, B,R. Hence you divide out all the ways you can organize the letters N, M, B, R.) ---stop here and check in with your teacher before continuing---- Then try 3.1 HW pg 109 # 1- 8