Question 1208901: How many numbers can be formed with five different digits from {0, 1, 2, 3, 4, 5, 6, 7} that must include the digits 2 and 5 ?
Answer by Edwin McCravy(20054) (Show Source):
You can put this solution on YOUR website!
We'll start out allowing 0 to be a first digit then we'll
count those and subtract.
There are 5 positions for the 5 digits.
Choose the position for the 2 in 5 ways.
Choose the position for the 5 in 4 ways.
There are three vacant positions remaining.
Choose a digit to go in the leftmost vacant position 6 ways.
Choose a digit to go in the middle vacant position 5 ways.
Choose a digit to go in the rightmost vacant position 4 ways.
That's (5)(4)(6)(5)(4) = 2400
We must now count the undesired ones, i.e., that begin with 0.
We can consider these to be 4-digit numbers with digits from
{1, 2, 3, 4, 5, 6, 7} that must include the digits 2 and 5.
There are 4 positions for the 4 digits.
Choose the position for the 2 in 4 ways.
Choose the position for the 5 in 3 ways.
There are two vacant positions remaining.
Choose a digit to go in the leftmost vacant position 5 ways.
Choose a digit to go in the rightmost vacant position 4 ways.
That's (4)(3)(5)(4) = 240
Subtracting, 2400-240 = 2160 <--ANSWER
Edwin
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