I am not doing the problem for you but perhaps I can give you an idea.
Let's see. I have to work with
M M M A A L S
If all letters were different the answer would be 7!, but they are not.
Let me try a simpler example
Say
A A B B
A B A B
A B B A
B A A B
B B A A
B A B A Hmm six of them, not 4!
Looks like permutations of 4 items divided by permutations of 2 As divided by permutations of 2 Bs
4!/(2!*2!) = 4*3*2 /( 2*2) = 6
Let's try it with 3 As and 1 B
A A A B
A A B A
A B A A
B A A A Is that 4!/(3!*1!) ?
4*3*2 /(3*2*1) = 4, sure enough.
For MAMMALS, find the munber of:
a) 7-letter words
b) 7-letter word beggining with MA
c) 4-letter words
i) all letters different
ii) 2 letters same,2 different
iii) 2 pairs same letters
iv) 3 letters same
1 answer