Difference between revisions of "1995 AJHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are <math>5*5*8*7= | + | Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are <math>5*5*8*7= \boxed{\text{(B)}\ 1400}</math> numbers. |
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1995|num-b=22|num-a=24}} | ||
Latest revision as of 02:23, 23 December 2012
Problem
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
Solution
Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are 
numbers.
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
