permutations with repititions

Question

In how many ways 10 different letters can be posed in 6 post boxes so that at most 4 boxes may be empty?

Comments

No I think not @vijay

it is distribution formula

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@Vijay, good. But what that's if letters are same rt?

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yes sir ...  my mistake ..

Answer 1

approach given by radha gogia is correct..
The answer can be calculated as:
= (6C₂)(2¹⁰) + (6C₃)(3¹⁰) + (6C₄)(4¹⁰) + (6C₅)(5¹⁰) + (6C₆)(6¹⁰)
= 135984906