Permutations And Combinations


The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet? 


Number of consonants = 21
                 Number of vowels = 5
A word contains two vowels and two consonants will be formed as
(i)  select 2 vowels out of 5
(ii)  select 2 consonants out of 21
(iii) arrange 2 + 2 = 4 letters to obtain different words
(iv) use fundamental principle of counting.
          (i) Number of ways in which 2 vowels can be selected out of 5 = straight C presuperscript 5 subscript 2
          (ii) Number of ways in which 2 consonants can be selected out of 21 = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/ at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/ line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
#1 /home/config_admin/public/ mkdir('/home/config_ad...', 493)
#2 /home/config_admin/public/ com_wiris_util_sys_Store->mkdirs()
#3 /home/config_admin/public/ com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...')
#4 /home/config_admin/public/ com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/ com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#6 {main}</pre>
           (iii) Number of permutations (to form different words) of the four letters = 4!
Hence, by fundamental principle of counting, the number of words formed.
                               equals straight C presuperscript 5 subscript 2 cross times straight C presuperscript 21 subscript 2 cross times 4 factorial space equals space fraction numerator 5 factorial over denominator 3 factorial space space 2 factorial end fraction cross times fraction numerator 21 space factorial over denominator 19 space factorial space space 2 space factorial end fraction cross times 4 factorial
                                = 10 x 210 x 24 = 50400


Sponsor Area

Some More Questions From Permutations and Combinations Chapter