AMC 10 · 2013 · #11

Grade 7 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

combinations-basicquadratic-equations convert-to-algebraguess-and-check ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 2 insights

Problem

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A student council has some number of members. Choosing a two-person welcoming committee can be done in exactly 10 ways. Students may serve on both committees, so the same members are available for the planning committee. Find how many different three-person planning committees can be chosen from the same council.

Givens: There are exactly 10 ways to choose the two-person welcoming committee.; The planning committee has three people.; A student may serve on both committees, so all members are available for both.

Unknowns: The number of different three-person planning committees.

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #2 Make a Systematic List, #6 Guess and Check

The council size is hidden. Name it with a variable, use the 10 ways to pick a pair to pin down how many members there are, then count the three-person committees for that many members.

Execute — Answer: A

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the number of members

Let n stand for the number of students on the council.
The welcoming committee is any group of 2 different members, and the order in which they are picked does not matter.

$$n = \text{number of council members}$$

💡 Giving the unknown count a name lets us turn the clue into an equation.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count 2-person groups

Pick a first member n ways and a second member n-1 ways, giving n(n-1) ordered picks.
Each pair is counted twice, once in each order, so divide by 2.
The number of two-person committees is n(n-1)/2.

$$\binom{n}{2}=\dfrac{n(n-1)}{2}$$

💡 Ordered picking counts every unordered pair twice, so halve it.

#6 Guess and Check 4.OA.B.4 Step 3

Solve for n

The clue says there are 10 such pairs, so n(n-1)/2 = 10, which gives n(n-1) = 20.
Look for two consecutive whole numbers whose product is 20: 5 and 4 work, so n = 5.

$$\dfrac{n(n-1)}{2}=10 \;\Rightarrow\; n(n-1)=20 \;\Rightarrow\; 5\cdot 4=20 \;\Rightarrow\; n=5$$

💡 Two consecutive numbers whose product is 20 must be 5 and 4.

#2 Make a Systematic List 7.SP.C.8 Step 4

Count 3-person groups

Now choose 3 members from the 5 on the council.
Pick 3 in order for 5 \cdot 4 \cdot 3 = 60 ways, but each group of 3 is counted 3 \cdot 2 \cdot 1 = 6 times, so divide.
There are 60/6 = 10 planning committees, so the answer is (A).

$$\binom{5}{3}=\dfrac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1}=\dfrac{60}{6}=10$$

💡 Divide the ordered picks by the number of ways to reorder the same group.

[1] #4 6.EE.B.6 Let n stand for the number of students on the council. The welcoming committee i

[2] #2 7.SP.C.8 Pick a first member n ways and a second member n-1 ways, giving n(n-1) ordered p

[3] #6 4.OA.B.4 The clue says there are 10 such pairs, so n(n-1)/2 = 10, which gives n(n-1) = 20

[4] #2 7.SP.C.8 Now choose 3 members from the 5 on the council. Pick 3 in order for 5 \cdot 4 \c

Review

Reasonableness: Check n = 5 against the first clue: choosing 2 of 5 gives 5 \cdot 4 / 2 = 10 pairs, which matches. Then choosing 3 of 5 also gives 10. A quick symmetry check confirms it: choosing 3 people to include is the same as choosing the 2 people to leave out, and there are 10 ways to leave out a pair. So the count is 10, answer (A).

Alternative: Skip the choose formula and label the members 1 through 5, then list every three-person group by hand. Careful listing produces exactly 10 groups, confirming the count without algebra.

CCSS standards used (min grade 7)

7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the two-person and three-person committees as organized groups (combinations))
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the council size n and expressing the pair count in terms of n)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Finding the consecutive factor pair 5 times 4 equals 20 to solve for n)

⭐ When a count is given to you, work backwards to find how many things there are, then count the new way.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Name the number of members

Count 2-person groups

Solve for n

Count 3-person groups

Review

CCSS standards used (min grade 7)

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