Difference between revisions of "2003 AMC 10A Problems/Problem 21"

Revision as of 23:15, 16 July 2014

Problem

Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?

$\mathrm{(A) \ } 22\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 27\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 729$

Solution 1

Let the ordered triplet $(x,y,z)$ represent the assortment of $x$ chocolate chip cookies, $y$ oatmeal cookies, and $z$ peanut butter cookies.

Using casework:

Pat selects $0$ chocolate chip cookies:

Pat needs to select $6-0=6$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(0,6,0); (0,5,1); (0,4,2); (0,3,3); (0,2,4); (0,1,5); (0,0,6)\} \rightarrow 7$ assortments.

Pat selects $1$ chocolate chip cookie:

Pat needs to select $6-1=5$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(1,5,0); (1,4,1); (1,3,2); (1,2,3); (1,1,4); (1,0,5) \} \rightarrow 6$ assortments.

Pat selects $2$ chocolate chip cookies:

Pat needs to select $6-2=4$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(2,4,0); (2,3,1); (2,2,2); (2,1,3); (2,0,4)\} \rightarrow 5$ assortments.

Pat selects $3$ chocolate chip cookies:

Pat needs to select $6-3=3$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(3,3,0); (3,2,1); (3,1,2); (3,0,3)\} \rightarrow 4$ assortments.

Pat selects $4$ chocolate chip cookies:

Pat needs to select $6-4=2$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(4,2,0); (4,1,1); (4,0,2)\} \rightarrow 3$ assortments.

Pat selects $5$ chocolate chip cookies:

Pat needs to select $6-5=1$ more cookies that are either oatmeal or peanut butter.

The assortments are: $\{(5,1,0); (5,0,1)\} \rightarrow 2$ assortments.

Pat selects $6$ chocolate chip cookies:

Pat needs to select $6-6=0$ more cookies that are either oatmeal or peanut butter.

The only assortment is: $\{(6,0,0)\} \rightarrow 1$ assortment.

The total number of assortments of cookies that can be collected is $7+6+5+4+3+2+1=28 \Rightarrow\boxed{\mathrm{(D)}\ 28}$

Solution 2

It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is $\binom{8}{2} = 28 \Rightarrow\boxed{\mathrm{(D)}\ 28}$

@@ Line 55: / Line 55: @@
 == Solution 2 ==
-It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is <math>\binom{8}{2} = 28 \Rightarrow \boxed{D}</math>
+It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is <math>\binom{8}{2} = 28 \Rightarrow\boxed{\mathrm{(D)}\ 28}</math>
 == See also==

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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Difference between revisions of "2003 AMC 10A Problems/Problem 21"

Revision as of 23:15, 16 July 2014

Contents

Problem

Solution 1

Solution 2

See also