Combination word problems help

Here are some carefully chosen combination word problems that will show you how to solve word problems involving combinations.

Use the combination formula shown below when the order does not matter

The combination word problems will show you how to do the followings:

Use the combination formula

Use the multiplication principle and the combination formula

Use the addition principle and the combination formula

Use the multiplication principle, addition principle, and combination formula

Word problem #1

There are 18 students in a classroom. How many different eleven-person students can be chosen to play in a soccer team?

Solution

The order in which students are listed once the students are chosen does not distinguish one student from another. You need the number of combinations of 18 potential students chosen 11 at a time.

Evaluate _nC_r with n = 18 and r = 11

₁₈C₁₁ =

18!
/
11!(18 — 11)!

₁₈C₁₁ =

18×17×16×15×14×13×12×11!
/
11!(7×6×5×4×3×2)

₁₈C₁₁ =

18×17×16×15×14×13×12
/
(7×6×5×4×3×2)

₁₈C₁₁ = 31824

There are 31824 different eleven-person students that can be chosen from a group of 18 students.

Word problem #2

For your biology report, you can choose to write about three of a list of four different animals. Find the number of combinations possible for your report.

Solution

The order in which you write about these 3 animals does not matter as long as you write about 3 animals.

Evaluate _nC_r with n = 4 and r = 3

There are 4 different ways you can choose 3 animals from a list of 4.

Word problem #3

A math teacher would like to test the usefulness of a new math game on 4 of the 10 students in the classroom. How many different ways can the teacher pick students?

Solution

The order in which the teacher picks students does not matter.

Evaluate _nC_r with n = 10 and r = 4

₁₀C₄ =

10×9×8×7×6!
/
4×3×2(6)!

There are 210 ways the teacher can pick students

Exercise 10

How many groups can be made from the word «house» if each group consists of 3 alphabets?

Exercise 11

Sarah has 8 colored pencils that are all unique. She wants to pick three colored pencils from her collection and give them to her younger sister. How many different combinations of colored pencils can Sarah make from 8 pencils?

Exercise 12

Alice has 6 chocolates. All of the chocolates are of different flavors. She wants to give two of her chocolates to her friend. How many different combinations of chocolates can Alice make from six chocolates?

Solution of exercise 1

How many different combinations of management can there be to fill the positions of president, vice-president and treasurer of a football club knowing that there are 12 eligible candidates?

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 2

How many different ways can the letters in the word «micro» be arranged if it always has to start with a vowel?

The words will begin with i or o followed by the remaining 4 letters taken from 4 by 4.

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 3

How many combinations can the seven colors of the rainbow be arranged into groups of three colors each?

The order of the elements does not matter.

The elements cannot be repeated.

$binom{7}{3} = frac {7!} {3! (7-3!)}$

$binom{7}{3} = frac {7!} {3! cdot 4!)} = 35$

Solution of exercise 4

How many different five-digit numbers can be formed with only odd numbered digits? How many of these numbers are greater than 70,000?

The order of the elements does matter.

The elements cannot be repeated.

n = 5 k = 5

The odd numbers greater than 70,000 have to begin with 7 or 9. Therefore:

Solution of exercise 5

How many games will take place in a league consisting of four teams? (Each team plays each other twice, once at each teams respective «home» location)

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 6

10 people exchange greetings at a business meeting. How many greetings are exchanged if everyone greets each other once?

The order of the elements does not matter.

The elements cannot be repeated.

$C_{10} ^ 2 = frac {10 cdot 9 } {2} = 45$

Solution of exercise 7

How many five-digit numbers can be formed with the digits 1, 2 and 3? How many of those numbers are even?

The order of the elements does matter.

The elements are repeated.

If the number is even it can only end in 2.

Solution of exercise 8

How many lottery tickets must be purchased to complete all possible combinations of six numbers, each with a possibility of being from 1 to 49?

The order of the elements does not matter.

The elements cannot be repeated.

$binom{49}{6} = frac {49!} {6! (49-6!)}$

Solution of exercise 9

How many ways can 11 players be positioned on a soccer team considering that the goalie cannot hold another position other than in goal?

Therefore, there are 10 players who can occupy 10 different positions.

The order of the elements does matter.

The elements cannot be repeated.

Solution of exercise 10

The word house has 5 alphabets. If each new word should have 3 alphabets, then we should use the following formula:

$binom{n}{k} = frac {n!} {k! (n-k!)}$ , where

Substitute the values in this example in the above formula:

$binom{5}{3} = frac {5!} {3! (5-3!)}$

Solution of exercise 11

Number of pencils Sarah have = 8

Number of pencils she wants to give to her younger sister = 3

We will use the binomial coefficients formula to determine the number of combinations:

$binom{n}{k} = frac {n!} {k! (n-k!)}$ , where

After substitution we will get the number of combinations:

$binom{8}{3} = frac {8!} {3! (8-3!)}$

$binom{8}{3} = frac {8!} {3! (5!)}$

$binom{8}{3} = frac {40320} {720}$

Hence, Sarah can make 56 combinations of 8 colored pencils given the fact that she can choose 3 at a time.

Solution of exercise 12

Number of chocolates Alice have = 6

Number of chocolates she wants to give to her friend= 2

We will use the binomial coefficients formula to determine the number of combinations:

$binom{n}{k} = frac {n!} {k! (n-k!)}$ , where

After substitution we will get the number of combinations:

$binom{6}{2} = frac {6!} {2! (6-2!)}$

$binom{6}{2} = frac {6!} {2! (4!)}$

$binom{6}{2} = frac {720} {48}$

Hence, Alice can make 15 combinations of 6 chocolates given the fact that she can choose 2.

Источник

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(1) Part 1 of 4 — How to Solve permutation & combination word problems, (2) Part 2 of 4 — How to Solve permutation & combination word problems, (3) Part 3 of 4 — How to Solve permutation & combination word problems, (4) Part 4 of 4 — How to Solve permutation & combination word problems

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Источник

Example 1 :

Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?

Solution :

Number of ways of selecting 7 players out of 14 players

= ¹⁴C₇

= 14! / (14 — 7)! 7!

= 14! / 7! 7!

= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7!)/7! 7!

= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ / 7!

= (14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ / (7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2)

= 3432

Example 2 :

There are 15 persons in a party and if each 2 of them shakes hands with each other, how many handshakes happen in the party?

Solution :

Before going to look into the solution of this problem, let us create a model.

By comparing the above results, we may conclude the formula to find number of hand shakes

Number of hand shakes = n (n — 1)/2

Here we divide n (n — 1) by 2, because of avoiding repetition.

Number of persons in a party = 15

Number of hand shakes can be made = 15 (15 — 1) / 2

= 15 (14)/2

= 15 (7)

= 105

Example 3 :

How many chords can be drawn through 20 points on a circle?

Solution :

20 points lie on the circle. By joining any two points on the circle, we may draw a chord.

Number of chords can be drawn = ²⁰C₂

= 20!/(20 — 2)! 2!

= 20! / 18! 2!

= (20 ⋅ 19) / 2

= 190

Hence the required number of chords can be drawn is 190.

Example 4 :

In a parking lot one hundred , one year old cars, are parked. Out of them five are to be chosen at random for to check its pollution devices. How many different set of five cars can be chosen?

Solution :

In the given question, we have a word «different set of five cars».

So we have to use the concept combination.

Number of ways of choosing 5 cars = ¹⁰⁰C₅

Hence the answer is ¹⁰⁰C_5.

Example 5 :

How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?

Solution :

Total number of boys	Number of boys to be selected	Ways
5	3	⁵C₃ = 10

Total number of girls	Number of girls to be selected	Ways
4	2	⁴C₂ = 6

Total number of transgenders	Number of transgenders to be selected	Ways
2	1	²C₁ = 2

Total number of ways = 10 ⋅ 6 ⋅ 2

= 120 ways

Hence the total number of ways is 120.

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Источник

I. Solve the following word problems. Show complete solution.

In how many ways can the letters of the word PERMUTATION be arranged?
Bob, Sue, Larry, Sally and Fred are waiting in line to buy concert tickets. In how many different ways can they stand in line?
A club consisting of eight members wishes to randomly select a president, vice-president and secretary, how many different arrangements are possible?
If six friends go to the movies.

a) How many different ways can they arrange themselves in a row of six seats?

b) Jodi, and Bek’ah are best friends, and must sit together. How many arrangements are possible with these 2 sitting next to each other?

c)Jodi and Bek’ah both like the same boy. Now they hate each other. How many ways can the six friends sit without Jodi and Bek’ah beside each other?

5. A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if

a) there are no restrictions?

b) one particular person must be chosen on the committee?

Pls explain how you the answer and give complete solutions.

Источник

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