SOLUTION: How many four-digit numbers can be formed from the digits 0, 1,2,3,4,5,and 6 if each digit can be used only once? (b) How many of these are odd numbers? (c)How many are greater t

Algebra ->  Probability-and-statistics -> SOLUTION: How many four-digit numbers can be formed from the digits 0, 1,2,3,4,5,and 6 if each digit can be used only once? (b) How many of these are odd numbers? (c)How many are greater t      Log On


   



Question 1201940: How many four-digit numbers can be formed from the digits 0, 1,2,3,4,5,and 6 if each digit can be used only once?
(b) How many of these are odd numbers?
(c)How many are greater than 330?

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(a) Total number of 4-digit numbers

There is one restriction: 0 can't be the first digit

6 choices for the first digit (1 to 6)
6 choices for the second (any of the remaining 6 digits, including 0)
5 choices for the third
4 choices for the fourth

ANSWER for (a): 6*6*5*4 = 720

(b) Number of 4-digit odd numbers

There are two restrictions: 0 can't be first, and the last digit must be odd. There are two cases to consider -- first digit odd or first digit even.

(b1) odd; first digit even

3 choices for first digit (2, 4, or 6)
3 choices for the last digit (1, 3, or 5)
5 choices for the second (any of the remaining 5 digits)
4 choices for the third
total: 3*3*5*4 = 180

(b2) odd, first digit odd

3 choices for the first digit (1, 3, or 5)
2 choices for the last (there are only 2 other odd digits)
5 choices for the second (any of the remaining 5 digits)
4 choices for the third
total: 3*2*5*4 = 120

ANSWER for (b): 180+120 = 300

(c) Number greater than 330

ANSWER: 720 (all 4-digit numbers are greater than 330...!)

NOTE: The "330" in the post is probably a typo; it was probably supposed to be a 4-digit number. So this was probably supposed to be a much more interesting problem -- but I am answering the question that was asked in the post....

NOTE: For a student just learning to work problems like this, it is a useful exercise to calculate how many of these 4-digit numbers are even. Since there are a total of 720 numbers, of which 300 are odd, there should be 420 that are even.

Let's see....

(d1) even, first digit even

3 choices for the first digit (2, 4, or 6)
3 choices for the last digit (any of the three remaining even digits, including 0)
5 choices for the second
4 choices for the third
total: 3*3*5*4 = 180

(d2) even, first digit odd

3 choices for the first digit (1, 3, or 5)
4 choices for the last digit (any of the 4 even digits)
5 choices for the second
4 choices for the third
total: 3*4*5*4 = 240

Total even: 180+240 = 420 CORRECT!