A) 150
B) 148
C) 149
D) None of these
Correct Answer: C
Solution :
Here we have \[1\ M,\ 4\ I,\ 4\ S\] and\[2P\]. Therefore total number of selections of one or more letters\[=(1+1)(4+1)(4+1)(2+1)-1=149\].You need to login to perform this action.
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