But out of these words few words will be same as the there are repeated letters in ALLAHABAD .
A – 4 times , L 2 times . Thus we have to ignore these repeated words . This can be done by dividing the solution by 4 ! & 2 !.
Thus number of 9 letters word formed using letters of ALLAHABAD is 362880 /( 4 ! * 2 !) = 7560 .
Now , let ’ s solve using formula . Number of 9 letters word formed using letters of 9 !/ 4 ! * 2 ! ) = 7560
In some cases , we need to fix some of the objects are certain positions while finding the permutation . For example if we have to arrange 5 students in 4 seats with a condition that 1 st seat is always occupied by a particular Student .
Numerical : For word INDEPENDENCE . In how many of these arrangements , ( i ) do words start with P ii ) do all vowels always occur together ( iii ) do vowels never occur together ( iv ) do words begin with I and end in P ?
Solution : There are 12 letters , of which N appears 3 times , E appears 4 times and D appears 2 times and the rest are all different .
( i ) If we fix the position of word P , we are left with 11 letters of which N appears 3 times , E appears 4 times and D appears 2 times and the rest are all different .
Therefore , the required numbers of words starting with P are 11 ! / ( 3 ! * 2 ! * 4 !) = 138600
( ii ) If all vowels occur together , then lets consider vowels { EEEEI } as one entity . Thus number of object now are {( EEEEI ), N , D , P , N , D , N , C ) that is 8 objects with B repeating 3 times & D repeating 2 times .
These 8 objects , in which there are 3Ns and 2 Ds , can be rearranged in 8 !/( 3 ! 2 !) ways .