ISEE Middle Level

Counting & Combinations

Fundamental counting principle, permutations, and combinations — how many ways to arrange or choose — excludes probability (see probability) and Venn diagram counting (see set-theory)

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Have you ever spent ages creating a custom character in a video game? 🎮 Imagine you can choose from 5 hairstyles, 4 shirts, and 3 pairs of shoes. How many totally unique characters can you make? You might be tempted to add the numbers, but the secret to unlocking all the possibilities is multiplying them! This is called the Fundamental Counting Principle, and it’s a superpower for the ISEE Quantitative Reasoning section.

Instead of drawing out every single outfit (which would take forever!), you just multiply: $5 \cdot 4 \cdot 3 = 60$ outfits! 🤯

Sometimes the ISEE will ask you to arrange things, like putting books on a shelf or letters in a password. If order matters (like a password), it's called a permutation. If order doesn't matter (like picking two friends to share a pizza), it's a combination. For the ISEE, the most important trick is to draw blank lines for each choice you need to make, fill in the number of options for each blank, and multiply them together. Let's get counting! 🚀

Key Formula

The Fundamental Counting Principle: Draw a blank line for each choice. Fill in the number of options for each, then multiply!

Total = Option_{1} \cdot Option_{2} \cdot Option_{3}

Practice Questions

4 practice questions for ISEE Middle Level

Q1 Medium

A deli allows customers to create their own sandwiches by choosing one type of bread, one type of meat, and one type of cheese. If there are 3 types of bread, 5 types of meat, and 4 types of cheese, how many different sandwich combinations are possible?

A 12

B 20

C 45

D 60

Show Solution

To find the total number of different sandwich combinations, we use the Fundamental Counting Principle. This principle states that if there are $n_{1}$ ways to do one thing, $n_{2}$ ways to do another, and $n_{3}$ ways to do a third, then there are $n_{1} \times n_{2} \times n_{3}$ ways to do all three.

Number of bread choices = 3
Number of meat choices = 5
Number of cheese choices = 4
Total combinations = $3 \times 5 \times 4 = 60$ .

Answer: D

Q2 Medium

From a collection of 10 distinct novels, how many different ways can a librarian arrange 3 of these novels side-by-side on a display shelf?

A 120

B 240

C 720

D 1000

Show Solution

This is a permutation problem because the order in which the books are arranged on the shelf matters. We are choosing 3 books from 10 and arranging them.

The formula for permutations is $P (n, k) = n! / (n - k)!$ , where $n$ is the total number of items, and $k$ is the number of items to arrange.
Here, $n = 10$ (total novels) and $k = 3$ (novels to arrange).
$P (10, 3) = 10! / (10 - 3)!$
$= 10! /7!$
$= 10 \times 9 \times 8 \times 7! /7!$
$= 10 \times 9 \times 8 = 720$ .
So, there are 720 different ways to arrange 3 novels from a collection of 10.

Answer: C

Q3 Medium

A small robotics club has 8 members. How many different ways can a committee of 3 members be chosen from the club?

A 24

B 56

C 112

D 336

Show Solution

This is a combination problem because the order in which the members are chosen for the committee does not matter. A committee consisting of members A, B, and C is the same as a committee consisting of B, C, and A.

The formula for combinations is $C (n, k) = n! / (k! \times (n - k)!)$ , where $n$ is the total number of items, and $k$ is the number of items to choose.
Here, $n = 8$ (total members) and $k = 3$ (members to choose for the committee).
$C (8, 3) = 8! / (3! \times (8 - 3)!)$
$= 8! / (3! \times 5!)$
$= (8 \times 7 \times 6 \times 5!) / ((3 \times 2 \times 1) \times 5!)$
$= (8 \times 7 \times 6) /6$
$= 8 \times 7 = 56$ .
So, there are 56 different ways to choose a committee of 3 members from the club.

Answer: B

Q4 Medium

How many different 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is allowed, and the number formed must be odd?

A 60

B 75

C 100

D 125

Show Solution

To form a 3-digit odd number using the digits 1, 2, 3, 4, 5 with repetition allowed, we consider the choices for each digit place:
1. Units digit (last digit): For the number to be odd, the units digit must be an odd number. From the given digits (1, 2, 3, 4, 5), the odd digits are 1, 3, and 5. So, there are $3$ choices for the units digit.
2. Tens digit (middle digit): Repetition is allowed, so any of the 5 given digits can be used for the tens digit. So, there are $5$ choices for the tens digit.
3. Hundreds digit (first digit): Repetition is allowed, so any of the 5 given digits can be used for the hundreds digit. So, there are $5$ choices for the hundreds digit.

Using the Fundamental Counting Principle, we multiply the number of choices for each digit to find the total number of possible odd 3-digit numbers:
Total number of odd 3-digit numbers = $(C h o i ces f or H u n d r e d s) \times (C h o i ces f or T e n s) \times (C h o i ces f or U ni t s)$
$= 5 \times 5 \times 3 = 75$ .

Answer: B

Tips & Strategies

Always check if repeats are allowed! If a question says 'no repeats' or 'without replacement', remember to subtract $1$ from your options for each new blank.
Draw it out! Actually draw blank lines on your scratch paper for each choice (like $_ \cdot _ \cdot _$ ). It makes the math so much easier to see.

Common Mistakes

Watch out for adding instead of multiplying! If you have 3 shirts and 4 pants, it's $3 \cdot 4 = 12$ outfits, NOT $3 + 4 = 7$ .
Don't forget to divide when order doesn't matter! If you are just picking a group of 2 people, picking 'Alex then Sam' is the same group as 'Sam then Alex'. You have to divide your total by $2 \cdot 1$ to remove the duplicates!

Frequently Asked Questions

How do I know if order matters in a question?

Ask yourself if swapping the items makes a new thing. Swapping numbers in a password (123 vs 321) makes a new password, so order matters! Swapping flavors in a bowl of ice cream (chocolate and vanilla) is still the same bowl, so order doesn't matter.

What if I get completely stuck on a counting question?

Remember that there is NO penalty for guessing on the ISEE! If you're stuck, try to eliminate an answer that seems way too small (like if the numbers were added instead of multiplied), pick your favorite letter, and move on.

Do I need to memorize big formulas for combinations?

Nope! For the ISEE, you just need to know the 'draw the blanks and multiply' trick. If order doesn't matter, just divide by the number of ways to arrange the items you picked. Keep it simple!

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