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Hello the question is above , I can't understand such type of questions, please help.

the answer is 479 but how does it come?

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As $A$ occurs $4$ times, we can choose A in $4+1$ ways

So, the number of ways of selection of no or more letters from the word: AAAABBCCCDEF will be $N=(4+1)(2+1)(3+1)(1+1)(1+1)(1+1)$

Clearly, this contains one combination of choosing no letter

So, the number of ways of selection of one or more letters will be $N-1$

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  • $\begingroup$ Thanks I got it, You always answer my questions lab bhattaharjee:) $\endgroup$ – dknight Oct 15 '13 at 16:12
  • $\begingroup$ @gaurav, my pleasure. Hope I can make the points clear $\endgroup$ – lab bhattacharjee Oct 15 '13 at 16:13
  • $\begingroup$ yo are very supporting, thanks again:) $\endgroup$ – dknight Oct 15 '13 at 16:17

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