Question 740459: How many permutations of the digits 0, 1, 2, . . . , 9 either
start with a 3 or end with a 7?
Found 2 solutions by lynnlo, ikleyn:
Answer by lynnlo(4176)   (Show Source):
Answer by ikleyn(53411)  (Show Source):
You can put this solution on YOUR website! .
How many permutations of the digits 0, 1, 2, . . . , 9 either
start with a 3 or end with a 7?
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Let P be the set of permutations of the digits 0, 1, 2, . . . , 9 that start with a 3.
This set has 9! permutations.
Let Q be the set of permutations of the digits 0, 1, 2, . . . , 9 that end with a 7.
This set has 9! permutations.
The sets P and Q have non-empty intersection set: it consist of all permutations that
start with a 3 and end with a 7.
This set (P ∩ Q) has 8! permutations.
So, the answer to the problem's question is this number
9! + 9! - 8! = 2*1*2*3*4*5*6*7*8*9 - 1*2*3*4*5*6*7*8 = 685440.
Solved.
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