SOLUTION: a) How many three digit numbers can be formed from the digits 2,3,4,5,6,7 and 8? No digit repeats in the same number. b) What is the probability of an even number? c) What is the

There are 7 digits from which 3 must be selected, and order matters:

(a) P(7, 3) = 7!/(7-3)! = 7!/4!  = 7*6*5 = 210



You are only supposed to post one problem per-post, but I'm in a good mood so I will give hint for (b):



Hint for (b):  how many numbers end with 2,4,6, or 8?   Figure out how many there are, then the probability is that number divided by the answer from (a).



If you look at the last digit of the 3 digit number, there are C(4,1) ways to select an even digit.  Once the ending digiit is known to be even, the other two digits can be even or odd.   So C(4,1) must be multiplied by P(6,2) which is the number of ways of arranging the other 2 digits out of the 6 remaining digits.



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For part (b), the value 1/6 is incorrect.  The correct answer is 4/7 (=120/210). Also notice that there are 4 even and 3 odd digits, so 4/7 are even.  Say you pick the ending digit first, that has a 4/7 chance of being even.  So simple!     







 

There are 7 digits.

Choose the 1st digit any of the 7 digits.  That's 7 ways.
Having chosen 1 digit, and since no digits can be repeated,
there remain only 6 digits to choose from.

Choose the 2nd digit any of the remaining 6 digits.  That's 6 ways.
Having chosen 2 digits, and since no digits can be repeated,
there remain only 5 digits to choose from. 

Choose the 3rd digit any of the remaining 5 digits. That's 5 ways.

That's 7×6×5 = 7P3 = 210

We first answer the question: "How many even three-digit numbers can
we have.  We choose the most restrictive thing first, which is the
3rd digit, for it must be even.

Choose the 3rd digit any of the 4 even digits 2,4,6,8.  That's 4 ways.
Having chosen 1 digit, and since no digits can be repeated,
there remain only 6 digits to choose from.

Choose the 1st digit any of the remaining 6 digits.  That's 6 ways.
Having chosen 2 digits, and since no digits can be repeated,
there remain only 5 digits to choose from.

Choose the 2nd digit any of the remaining 5 digits.  That's 5 ways.

So that's 4×6×5 = 120 ways

So the probability of an even number is 120 out of 720 or
120/720 which reduces to 1/6.

The first digit is either 5, 6 or 7, but we must take each of those
as a separate case because with each, the number of choices for the
2nd digit changes.

Case 1.  the first digit is 5.

Choose the first digit as 5. That's 1 way.
Choose the 2nd digit any of the 3 digits 6,7,8.  That's 3 ways.
Having chosen 2 digits, and since no digits can be repeated,
there remain only 5 digits to choose from for the 3rd digit.

That's 1×3×5 = 15 ways for case 1.

Case 2.  the first digit is 6.

Choose the first digit as 6.  That's 1 way.
Choose the 2nd digit as any of the 6 digits 2,3,4,5,7,8.  That's 6 ways.
Having chosen 2 digits, and since no digits can be repeated,
there remain only 5 digits to choose from for the 3rd digit.

That's 1×6×5 = 30 ways for case 2. 

Case 3.  the first digit is 7.

Choose the first digit as 7.  That's 1 way
Choose the 2nd digit any of the 5 digits 2,3,4,5,6.  That's 5 ways.
Having chosen 2 digits, and since no digits can be repeated,
there remain only 5 digits to choose from. That's 5 ways.

That's 1×5×5 = 25 ways for case 3.

So for all three cases, that's a grand total of 15+30+25 = 70 ways

So the probability of a number bettween 550 and 770 is 
70 ways out of 210 or 70/210 which reduces to 1/3.

Edwin