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| Author: | iim [ Fri Sep 28, 2007 10:26 am ] |
| Post subject: | A question in permutation combination |
How many five letter words comprising 2 vowels and 3 consonants be formed using the alphabet of the English language if all the five letters are to be different? |
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| Author: | ram [ Sun Oct 07, 2007 12:42 pm ] |
| Post subject: | |
Number of ways of selecting 2 vowels is 5C2. Number of ways of selecting 3 consonants is 21C3. Then arrange these 5 letters in 5! ways ergo 5C2 * 21C3 * 5! |
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| Author: | ag [ Mon Oct 08, 2007 11:49 am ] |
| Post subject: | |
Why not 5P2 * 21P3? I computed the values of both 5C2 * 21C3 * 5! and 5P2 * 21P3. I am getting different answers. What is the correct answer? |
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| Author: | iim [ Tue Oct 09, 2007 12:00 pm ] |
| Post subject: | 5P2 and 21P3 include arrangement within respective categorie |
When you take 5P2 instead of 5C2 while selecting vowels, you have actually re arranged vowels between them. i.e, it will count an arrangement of AE as different from EA. However, when you take it as 5C2, it will consider AE as the same as EA. Or in other words, 5C2 looks at having two vowels and not the order. The same goes for 21P3. So, when you multiply 5P2 and 21P3 what you have actually found out is the number of words that can be formed with the first two letters as two different vowels and the last 3 letters as 3 different consonants. You did not consider other possibilities. For eg. of the 5 letters are A, B, C, D and E then you have counted cases such as AEBCD or EADBC. But have missed out cases such as ABCDE or DCBEA. Hope this explanation helps. =============== 2IIM - CAT 2008 Long term batches begin Oct 14, 07 |
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