combi

[tushar7]

Find the probability that a five letter word that uses letters from the set $\{A.B.C\}$ contains at least one of each letter ?

my answer is $\frac{6.3^2}{3^5}$ ....... am i right??

[HandaramTheGreat]

$\frac{\binom{3}{1}\cdot\frac{5!}{3!}+\binom{3}{2}\cdot\frac{5!}{2!\cdot2!}}{3^5}$

[HandaramTheGreat]

গুণের জন্য \cdot দাও, তোমারটা দশমিক মনে হয়...

[tushar7]

$\frac{\binom{3}{1}\cdot\frac{5!}{3!}+\binom{3}{2}\cdot\frac{5!}{2!\cdot2!}}{3^5}$

can you explain it a little bit

[HandaramTheGreat]

i'm explaining it's numerator...
there should be one of each letter A, B, C... then we need 2 more letters... these 2 letters can be different or same...

one kind of letter(for 2 same letters) can be chosen in $\binom{3}{1}$ ways... then we can arrange 5 letters in $\frac{5!}{3!}$ ways, as there are 3 same letters in each choice(considering previous A, B, C also)...

2 different letters can be chosen in $\binom{3}{2}$ ways... then we can arrange 5 letters in $\frac{5!}{2!\cdot2!}$ ways, as there are 2 same letters twice in each choice...

[tushar7]

thanks jannat apu . i was too confused .

[Labib]

Bonji, tomar ki mone hoi na tumi aktu complicated hishab korteso!!
btw tomar oi soln purna sonkhai ashtese to??? ami calc dia fraction paitesi kan??

[Labib]

btw, tushar er songe amar sol'n mile! :p

[HandaramTheGreat]

tomar ki mone hoi na tumi aktu complicated hishab korteso!!

you should consider repetition... i solved it with 2 steps for counting permutations with repetition...

ami calc dia fraction paitesi kan??

probability can be fraction...

[Zzzz]

Tushar, please explain how you have got $\frac{6\cdot3^2}{3^5}$ ? May be you have done some mistakes

আর আমার সামান্য confusion আছে। শব্দগুলা কি খালি A,B,C দিয়ে তৈরি নাকি এর বাইরের অক্ষর ব্যবহার করা যাবে? যদি না যায় তাহলে জান্নাতের সমাধান ঠিক আছে।