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Peiges |
Latest page update: made by Peiges
, Apr 26 2007, 11:46 PM EDT
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Edited by Peiges
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Keyword tags:
distinct
Identical Elements
Permutations
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| Started By | Thread Subject | Replies | Last Post | ||
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| Anonymous | xfcgf | 0 | Apr 10 2012, 2:34 PM EDT by Anonymous | ||
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| Anonymous | PERMUTATIONS | 2 | Feb 14 2012, 11:47 AM EST by Anonymous | ||
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Thread started: Feb 3 2012, 8:51 AM EST
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How many 2 letter words can be formed from letters of the word " SPEECH" ?
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| Anonymous | Example 2 done differently: pls. tell if this correct | 0 | Dec 25 2011, 1:33 PM EST by Anonymous | ||
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Thread started: Dec 25 2011, 1:33 PM EST
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I was using the concept that you explained where we divide by the common permutations
Case 1: First digit is 1 and last digit is 6 1st and last positions are taken by 1 and 6. Now we have positions 2 to 6 left to be filled by 2,2,3,5,5 = total unique permutations/Common Permutation cases = 5!/(2! x 2!) Comments: there are two 2s and two 5s Case 2: First digit is 1 and last digit is 2 1st and last positions are taken by 1 and one of 2s. Now we have positions 2 to 6 left to be filled by 2,3,5,5,6 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in position 7, so only one 2 was left for middle. 5 repeats twice Case 3: First digit is 2 and last digit is 6 1st and last positions are taken by one of 2 and 6. Now we have positions 2 to 6 left to be filled by 1,2,3,5,5 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in position 1, so only one 2 was left for middle. 5 repeats twice Case 4: First digit is 2 and last digit is 2 1st and last positions are taken by 2 and 2. Now we have positions 2 to 6 left to be filled by 1,3,5,5,6 = total unique permutations/Common Permutation cases = 5!/(2!) Comments: there are two 5s. One of the 2 is tied up in positions 1 and 7. 5 repeats twice This will result in 60 + 120 + 120 + 120 = 420 cases. Please let me know if this understanding correct. |
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