1) if a monkey is given 11 letters that can form the word

1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

Hey, this is same as your previous answer:
1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

Your previous answer was also correct.

oh right ! too late here in germany you can approach this problem with combination- as well as with permutation-reasoning...

Can you guys explain 1?

divide by 2! and 2! because there are 2 b`s and 2 `s. for example probabilty with the first b in 4th position and the second b in 6th position is 1 way and when you change the b`s you get a second way, but it is the same word. so you divide 2.

the number 2) was not so easy, but I got 4!/11! and 1/48^6

rc1979, christoph explained it well, there are 2 letters B (le'ts call it B1 and B2 for example)
Usually you have to work with permutation concerning words problems because order normally matters, in that case there will be no difference if you choose B1 or B2, it will be the same word whereas each B is considered as a specific letter. So you'll have twice more chances with B and it's the same idea for I.

finally, normally for a word with 11 different litters the prob will be 1/11! but since you have 2 repetitions of 2 letters, it will be 1*2*2/11!

Hope it will help you to figure it out...

it is a permutation not a combiation problem. so it is
=1/(49P6)