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.