SOLUTION: How many 3-digit numbers can be formed from the digits 0,1,2,3,4,5,6 if each digit can be used only once? (ii) How many of these are odd numbers ? (iii) How many are greate

Algebra ->  Permutations -> SOLUTION: How many 3-digit numbers can be formed from the digits 0,1,2,3,4,5,6 if each digit can be used only once? (ii) How many of these are odd numbers ? (iii) How many are greate      Log On


   



Question 1014286: How many 3-digit numbers can be formed from the digits
0,1,2,3,4,5,6 if each digit can be used only once?
(ii) How many of these are odd numbers ?
(iii) How many are greater than 330 ?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

How many 3-digit numbers can be formed from the digits
0,1,2,3,4,5,6 if each digit can be used only once?
There are 7 unchosen digits
Choose the first digit any of 6 ways other than 0. 
That leaves 6 unchosen digits
Choose the second digit any of 6 ways. Can choose 0.
That leaves 5 unchosen digits.
Choose the third digit any of 5 ways.

Answer: (6)(6)(5) = 180 ways

(ii) How many of these are odd numbers ?
There are 7 unchosen digits.
Choose the last digit any of 3 ways, 1,3,or 5
That leaves 6 unchosen digits, including 0.
Choose the first digit any of 5 ways. (Can't choose 0).
That leaves 5 unchosen digits including 0.
Choose the middle digit any of these 5 unchosen ways.
Answer: (3)(5)(5) = 75 ways

(iii) How many are greater than 330 ?
Case 1:  All those between 330 and 400

Choose the first digit 1 way (as 3)
That leaves 6 unchosen digits.
Choose the second digit any of 3 ways, 4,5, or 6
That leaves 5 unchosen digits
Choose the third digit any of these 5 ways.

Answer case 1: (1)(3)(5) = 15 ways.

Case 2: Those greater than 400

Choose the first digit any of 3 ways, 4,5, or 6
That leaves 6 remaining unchosen digits.
Choose the second digit any of these 6 remaining unchosen digits.
That leaves 5 remaining unchosen digits.
Choose the third digit any of these 5 remaining unchosen digits.

Answer to case 2:  (3)(6)(5) = 90 ways.

Total for part (iii):  15+90 = 105 ways. 

Edwin