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Difference between revisions of "1996 AJHSME Problems/Problem 21"
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There are <math>2</math> choices for the even number.
 
There are <math>2</math> choices for the even number.
  
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There are <math>4</math> choices for the first odd number.  There are <math>3</math> choices for the last odd number.  But the order of picking these numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of <math>2</math>.  Thus, we have <math>\frac{4\cdot 3}{2} = 6</math> choices for a pair of odd numbers.
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There are <math>4</math> choices for the first odd number.  There are <math>3</math> choices for the last odd number.  But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of <math>2</math>.  Thus, we have <math>\frac{4\cdot 3}{2} = 6</math> choices for a pair of odd numbers.
  
 
In total, there are <math>2</math> choices for an even number, and <math>6</math> choices for the odd numbers, giving a total of <math>2\cdot 6 = 12</math> possible choices for a 3-element set that has an even sum.  This is option <math>\boxed{D}</math>.
 
In total, there are <math>2</math> choices for an even number, and <math>6</math> choices for the odd numbers, giving a total of <math>2\cdot 6 = 12</math> possible choices for a 3-element set that has an even sum.  This is option <math>\boxed{D}</math>.
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==See Also==
 
==See Also==

Latest revision as of 18:43, 26 May 2021

Problem

How many subsets containing three different numbers can be selected from the set \[\{ 89,95,99,132, 166,173 \}\] so that the sum of the three numbers is even?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution

To have an even sum with three numbers, we must add either $E+O+O$, or $E + E + E$, where $O$ represents an odd number, and $E$ represents an even number.

Since there are not three even numbers in the given set, $E+E+E$ is impossible. Thus, we must choose two odd numbers, and one even number.

There are $2$ choices for the even number.

There are $4$ choices for the first odd number. There are $3$ choices for the last odd number. But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$. Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.

In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum. This is option $\boxed{D}$.

See Also

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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