Question

how many different words can be formed by arranging letters of the word Ahmedabad such that(a) there is no restriction(b) with H and M hm and at the extreme (c)all consonants are together

Answer 1

may be 2 words with no restriction of h and m and one word where consonants are together

Answer 2

Answer:

(a) there are no restriction = $\frac{9!}{2!3!}$ = 30240

(b) H and M at the extreme = 840 words

(c) all consonants together = 480 words

Step-by-step explanation:

A H M E D A B A D ----------9 letters

A - repeated 3 times

D - repeated 2 times

the number of permutations of n objects, taken all at a time where there are p objects of similar type and q objects of another similar type

the arrangement is given as $\frac{n!}{p! q!}$

(a) there are no restriction = $\frac{9!}{2!3!}$ = 30240

30240 words can be formed without any restriction

(b) H and M at the extreme

       H _ _ _ _ _ _ _ M  _____ there will 7 letters in between H and M

out of this A is repeated 3 times D is repeated 2 times

then the number of possible arrangements = $\frac{2(7!)}{3!2!}$ = 840

840 words can be formed when H and M are at the extreme

(c) all consonants together

consonants are  H M D B D

these consonants can arrange themselves in $\frac{5!}{2!}$  = 60 ways

the vowels are A A A E

these vowels can arrange themselves in $\frac{4!}{3!}$ = 4 ways

consonants and vowels can be arranged as vowel consonant or consonant vowel =  $2!$ =  2 ways

number of arrangements with all consonants together =  $\frac{5!}{2!}$  × $\frac{4!}{3!}$ × $2!$

          = 60 × 4 × 2 = 480 words