Permutations And Combinations

Question

# How many different words, with or without meaning can be formed, by using the letters of the word ‘HARYANA’? Also, find as to:(a) how many of these begin with H and end with N?(b) in how many of these H and N are together?

Solution

Total letters in 'HARYANA' = 7
H → 1, A → 3, R → 1, Y → 1, N → 1.
n = 7, p = 3
∴           Number of arrangements =
(a)    The word begins with H and ends with N.
Fix the first place with H.
The number of permutations =                                                           ...(i)
Fix the last place with N.
The number of permutations =                                                            ...(ii)
Now, we are left with 5 places and 3(A) + 1 (R) + 1 (Y)
Number of arrangements =                              ...(iii)
∴     By fundamental principle of counting, the number of arrangements
= 1 x 1 x 20 = 20.
(b)  When H and N are together.
Tie the two.
Number of arrangements =                                                 ...(i)
Mix the tied bundle with the remaining.
[3 (A) + 1 (R) + 1(Y)] + 1 = 6 letters

Number of permutations =                          ...(ii)
∴    From (i) and (ii), applying fundamental principle of counting, we get the number of arrangements when H and N are together
= 2 x 120 = 240.