There are also two types of combinations (remember the order does not matter now):
Note: The primary difference between combinations and permutations is that in the case of combinations, the order does not matter but in permutations, the order does matter.
In this case, the elements of a set cannot be repeated, for example: in a company where there work 10 people they take the decision of forming a directive composed of 3 people(one person will be president, another one as secretary, and last one as treasurer), in this case, we would have a combination without repetition, because a person cannot be chosen twice.
Formula without repetition is given by
\(C(n,r)={ }^n C_r=\left(\begin{array}{l}where \(n\) is the number of things to choose from, and we choose \(r\) of them, no repetition, the order doesn’t matter.
Example 1: 2 girls will go to a party, if between the two, they have 4 pairs of fancy shoes, define the combination of shoes these two girls can wear.
Solution:
In this case \(n=4\) and \(r=2\)
\(Example 2: Father asks his son to choose 4 items from the table. If the table has 18 items to choose, how many different answers could the son give?
Solution:
Given,
\(
\begin{aligned}
& r=4 \text { (item sub-set) } \\
& n=18 \text { (larger item) }
\end{aligned}
\)
Therefore, simply: find ” 18 Choose 4 “
We know that, Combination \(=C(n, r)=n ! / r !(n-r)\) !
\(
\frac{18 !}{4 !(18-4) !}=\frac{18 !}{14 ! \times 4 !}
\)
\(
=3,060 \text { possible answers. }
\)
Example 3: From a class of 30 students, 4 are to be chosen for the competition. In how many ways can they be chosen?
Solution:
Total Students \(=n=30\)
Number of students to be chosen \(=r=4\)
Hence, the total number of ways 4 students out of 30 can be chosen is,
\(
\begin{aligned}
{ }^{30} C_4 & =(30 \times 29 \times 28) /(4 \times 3 \times 2 \times 1) \\
& =24360 / 24 \\
& =1015 \text { ways }
\end{aligned}
\)
Example 4: Nitin has 5 friends. In how many ways can he invite one or more of them to his party.
Solution:
Nitin may invite (i) one of them (ii) two of them (iii) three of them (iv) four of them \((v)\) all of them
and this can be done in \({ }^5 C_1,{ }^5 C_2,{ }^5 C_3,{ }^5 C_4,{ }^5 C_5\) ways
Therefore, The total number of ways \(={ }^5 C_1+{ }^5 C_2+{ }^5 C_3+{ }^5 C_4+{ }^5 C_5\) \(=5 ! /(1 ! 4 !)+5 ! /(2 ! 3 !)+5 ! /(4 ! 1 !)+5 ! /(5 ! 0!)\)
\(
\begin{aligned}
& =5+10+10+5+1 \\
& =31 \text { ways }
\end{aligned}
\)
Example 5: Find the number of diagonals that can be drawn by joining the angular points of an octagon.
Solution:
A diagonal is made by joining any two angular points.
There are 8 vertices or angular points in an octagon
Therefore, the Number of straight lines formed \(={ }^8 C_2=8 ! /(2 ! 6 !)\)
\(
\begin{aligned}
& =8 \times 7 /(2 \times 1) \\
& =56 / 2 \\
& =28
\end{aligned}
\)
which also includes the 8 sides of the octagon
Therefore, Number of diagonal \(=28\) – sides of octagon
\(
\begin{aligned}
& =28-8 \\
& =20 \text { diagonals }
\end{aligned}
\)
Example 6: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 7 blue balls, if each selection consists of 3 balls of each color.
Solution:
Number of Red balls \(=6\)
Number of white balls \(=5\)
Number of Blue balls = 7
Total number of balls to be selected \(=9\)
Hence, the required number of ways of selecting 9 balls from 6 red, 5 white, 7 blue balls consisting of 3 balls of each colour.
\(
\begin{aligned}
& ={ }^6 C_3 \times{ }^5 C_3 \times{ }^7 C_3 \\
& =6 ! /(3 ! 3 !) \times 5 ! /(3 ! 2 !) \times 7 ! /(3 ! 4 !) \\
& =20 \times 10 \times 35 \\
& =7000 \text { ways }
\end{aligned}
\)
Example 7: A group of 3 lawn tennis players S, T, U. A team consisting of 2 players is to be formed. In how many ways can we do so?
Solution:
In a combination problem, we know that the order of arrangement or selection does not matter.
Thus \(\mathrm{ST}=\mathrm{TS}, \mathrm{TU}=\mathrm{UT}\), and \(\mathrm{SU}=\mathrm{US}\).
Thus we have 3 ways of team selection.
By combination formula we have-
\(
\begin{aligned}
& { }^3 \mathrm{C}_2=3 ! / 2 !(3-2) ! \\
& =(3.2 \cdot 1) /(2 \cdot 1.1)=3
\end{aligned}
\)
Example 8: Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} having 3 elements.
Solution:
The set given here has 10 elements. We need to form subsets of 3 elements in any order. If we select \(\{1,2,3\}\) as the first subset then it is the same as \(\{3,2,1\}\). Hence, we will use the formula of combination here.
Therefore, the number of subsets having 3 elements \(={ }^{10} \mathrm{C}_3\)
\(
\begin{aligned}
& =10 ! /(10-3) ! 3 ! \\
& =10.9 .87 ! / 7 ! \cdot 3 ! \\
& =10.9 .8 / 3.2 \\
& =120 \text { ways. }
\end{aligned}
\)
Example 9: Find the number of 4-letter combinations which can be made from the letters of the word DRIVEN.
Solution:
\(Example 10: Out of a group of 5 people, a pair needs to be formed. Find the number of possible combinations.
Solution:
\(Example 11: In a party of 10 people, each person shakes hands with every other person. How many possible combinations of handshakes can be made?
Solution:
Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X, Y} will be counted. Here we consider the combination of any two people who shake their hand.
Thus, we need to find the number of ways in which 2 people can be selected out of 10.
\(Example 12: A class has 25 students. For a school event, 10 students need to be chosen from this class. 4 of the students of the class decide that either four of them will participate in the event, or none of them will participate. What are the possible combination of 10 students?
Solution:
Given the specified constraint, we divide the set of all possible selections of 10 students into two groups:
The selections include the 4 students; we already have 4 students – we need to select 6 more students out of the remaining 21 students. This can be done in \({ }^{21} \mathrm{C}_6\) ways. Thus, the number of possible selections that include the 4 students is \({ }^{21} \mathrm{C}_6\).
The selections do not include the 4 students; now we need to select 10 students out of the remaining 21. This can be done in \({ }^{21} \mathrm{C}_{10}\) ways. Thus, the number of possible selections that do not include the 4 students is \({ }^{21} \mathrm{C}_{10}\).
The total number of possible combinations under the specified constraint is \({ }^{21} \mathrm{C}_6+{ }^{21} \mathrm{C}_{10}\).
Example 13: How many diagonals are there in a polygon with \(n\) sides?
Solution:
A polygon of \(n\) sides has \(n\) vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon. Number of line segments obtained by joining the vertices of an \(\mathrm{n}\) sided polygon taken two at a time \(=\) Number of ways of selecting 2 out of \({n}={n}_{\mathrm{C}_2}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
Out of these lines, \(n\) lines are the sides of the polygon.
Therefore, Number of diagonals of the polygon \(=\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}\).
Example 14: 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
\(
\begin{gathered}
={ }^{14} C_7 \\
=14 ! /(14-7) ! 7 != 3432
\end{gathered}
\)
Example 15: 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 \(={ }^{20} C_2\)
\(
\begin{aligned}
& =20 ! /(20-2) ! 2 ! \\
& =20 ! / 18 ! 2 ! \\
& =(20 \cdot 19) / 2 \\
& =190
\end{aligned}
\)
Hence the required number of chords that can be drawn is 190.
Example 16: In a parking lot one hundred, one-year-old cars, are parked. Out of them five are to be chosen at random to check their pollution devices. How many different sets of five cars can be chosen?
Solution:
In the given question, it is given “different set of five cars”.
So we have to use the concept combination.
\(
\text { Number of ways of choosing } 5 \text { cars }={ }^{100} C_5
\)
Hence the answer is \({ }^{100} C_5\).
Example 17: 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 | \({ }^5 C_3=10\) |
| Total number of girls | Number of girls to be selected | Ways |
| 4 | 2 | \({ }^4 C_2=6\) |
| Total number of transgenders | Number of transgenders to be selected | Ways |
| 2 | 1 | \({ }^2 c_1=2\) |
Total number of ways = 10 ⋅ 6 ⋅ 2 = 120 ways. Hence the total number of ways is 120.
Example 18: Suppose we have an office of 5 women and 6 men and we have to select a committee of 4 people. In how many ways can we select 2 men and 2 women?
Solution:
In this case, we have to find two different combinations and then multiply them. Therefore, we want to calculate \(\left({ }^5 C_2\right)\left({ }^6 C_2\right)\). We can calculate these combinations separately:
\(
\begin{aligned}
& { }^5 C_2=\frac{5 !}{(5-2) ! 2 !} \\
& =\frac{5 !}{(3) ! 2 !} \\
& =\frac{5 \times 4 \times 3 !}{(3) ! 2 !} \\
& =\frac{5 \times 4}{2 !}=10 \\
& { }^6 C_2=\frac{6 !}{(6-2) ! 2 !} \\
& =\frac{6 !}{(4) ! 2 !} \\
& =\frac{6 \times 5 \times 4 !}{(4) ! 2 !} \\
& =\frac{6 \times 5}{2 !}=15 \\
&
\end{aligned}
\)
Therefore, we have \(\left({ }^5 C_2\right)\left({ }^6 C_2\right)=10 \times 15=150\).
Example 19: How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
From the given question, we come to know that any three points are not collinear.
By selecting any three points out of 15 points, we draw a triangle.
Number of ways to draw a triangle \(={ }^{15} C_3\)
\(
\begin{gathered}
=(15 \cdot 14 \cdot 13) /(3 \cdot 2 \cdot 1) =455
\end{gathered}
\)
Example 20: A committee of 7 members is to be chosen from 6 artists, 4 singers, and 5 writers. In how many ways can this be done if in the committee there must be at least one member from each group and at least 3 artists?
Solution:
For the given condition, possible ways to select members for a committee of 7 members.
\(
\begin{aligned}
& (3 \mathrm{~A}, 3 S, 1 \mathrm{~W}) —–> 6 C 3 \cdot 4 C 3 \cdot 5 C 1=20 \cdot 4 \cdot 5=400 \\
& (3 \mathrm{~A}, 15,3 \mathrm{~W}) —–> 6 C 3 \cdot 4 C 1 \cdot 3 C 1=20 \cdot 4 \cdot 10=800 \\
& (3 \mathrm{~A}, 25,2 \mathrm{~W}) —–> 6 C 3 \cdot 4 C 2 \cdot 5 C 2=20 \cdot 6 \cdot 10=1200 \\
& (4 \mathrm{~A}, 2 \mathrm{C}, 1 \mathrm{~W}) —–> 6 C 4 \cdot 4 C 2 \cdot 5 C 1=15 \cdot 6 \cdot 5=450 \\
& (4 \mathrm{~A}, 15,2 \mathrm{~W}) —–> 6 C 4 \cdot 4 C 1 \cdot 5 C 2=15 \cdot 4 \cdot 10=600 \\
& (5 \mathrm{~A}, 15,1 \mathrm{~W}) —–> 6 C 5 \cdot 4 C 1 \cdot 5 C 1=6 \cdot 4 \cdot 5=120
\end{aligned}
\)
Thus, the total no. of ways is
\(
\begin{gathered}
=400+800+1200+450+600+120 =3570
\end{gathered}
\)
Example 21: Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Find the number of trials in which the room can be lighted.
Solution:
3 bulbs are defective out of 5. There are two bulb points in the dark room.
One bulb (or two bulbs) in good condition is enough to light the room.
Since there are two bulb points, we have to select 2 out of 5 bulbs.
No. of ways of selecting 2 bulbs out of 5 is \({ }^5 C_2=10\)
(It includes selecting two good bulbs, two defective bulbs, one good bulb, and one defective bulb. So, in these 10 ways, the room may be lighted or may not be lighted)
The number of ways of selecting 2 defective bulbs out of 3 is \({ }^3 C_2=3\)
(It includes selecting only two defective bulbs. So, in these 3 ways, the room can not be lighted)
The number of ways in which the room can be lighted is = 10 – 3 = 7
Example 22: The supreme court has given a 6 to 3 decision upholding a lower court. Find the number of ways it can give a majority decision reversing the lower court.
Solution:
Upholding a lower court means, supporting it for its decision.
Reversing a lower court means, opposing it for its decision.
In a total of 9 cases \((6+3=9)\), it may give 5 or 6 or 7 or 8, or 9 decisions reversing the lower court. And it can not be 4 or less than 4. Because the majority of 9 is 5 or more.
The possible combinations in which it can give a majority decision reversing the lower court are
\(
\begin{aligned}
& 5 \text { out of } 9 \cdots> 9 C 5=126 \\
& 6 \text { out of } 9 \cdots> 9 C 6=84 \\
& 7 \text { out of } 9 \cdots> 9 C 7=36 \\
& 8 \text { out of } 9 \cdots> 9 C 8=9 \\
& 9 \text { out of } 9 \cdots> 9 C 9=1
\end{aligned}
\)
Thus, the total number of ways is
\(
\begin{gathered}
=126+84+36+9+1=256
\end{gathered}
\)
Example 23: A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
Number of white balls \(=2\)
Number of black balls \(=3\)
Number of red balls \(=4\)
Number of non-black balls \(=2+4=6\)
| Number of black balls to be drawn | Number of red balls to be drawn | Total balls to be drawn |
| \({ }^3 c_1\) | \({ }^6 C_2\) | 3 |
| \({ }^3 C_2\) | \({ }^6 C_1\) | 3 |
| \({ }^3 C_3\) | \({ }^6 C_0\) | 3 |
Number of ways
\(
\begin{aligned}
& =\left({ }^3 c_1 \cdot{ }^6 c_2\right)+\left({ }^3 c_2 \cdot{ }^6 c_1\right)+\left({ }^3 c_3 \cdot{ }^6 c_0\right) \\
& =(3 \cdot 15)+(3 \cdot 6)+(1 \cdot 1) \\
& =45+18+1 \\
& =64 \\
&
\end{aligned}
\)
Example 24: In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women?
Solution:
Choose 5 men out of 9 men \(={ }^9 C_5\) ways \(=126\) ways
Choose 3 women out of 12 women \(=12\) C3 ways \(=220\) ways
Total number of ways \(=(126 \times 220)=27720\) ways
The committee can be chosen in 27720 ways.
Example 25: Six friends want to play enough games of chess to be sure everyone plays everyone else. How many games will they have to play?
Solution:
There are 6 players to be taken 2 at a time. Using the formula:
\(
{ }^6 C_2=\frac{6 \times 5}{2 \times 1}=15
\)
They will need to play 15 games.
Example 26: In a lottery, each ticket has 5 one-digit numbers 0-9 on it. You win if your ticket has the digits in any order. What are your chances of winning?
Solution:
There are 10 digits to be taken 5 at a time. Using the formula:
\(
C(10,5)={ }^{10} C_5=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 !}=252
\)
The chances of winning are 1 out of 252.
Example 27: At a party, every person shakes hand with each other person. If there was a total of 26 handshakes done at the party, how many persons were present at the party?
Solution:
Suppose there are \(n\) persons present in a party and every person shakes hand with every other person in the party. Then, total number of handshakes will be counted as \(={ }^n C_2=n(n-1) / 2\)
\(
\begin{aligned}
& n(n-1) / 2=29 \\
& n(n-1)=26 \times 2 \\
& n(n-1)=52 \\
& n=8 \text { Persons }
\end{aligned}
\)
Example 28: A teacher gives an exam with 10 problems to choose from. The instructions read that only 8 of them need to be completed. In how many different ways can you choose which problems to complete on the exam?
Solution:
In this case, the total number of problems given is 10. The size of our combination will be 8 because that’s how many problems you are supposed to complete. We can use the combination formula
\(
{ }^{10} C_8=\frac{10 !}{8 !(10-8) !}=\frac{10 \times 9 \times 8 !}{8 ! \times 2 !}=\frac{10 \times 9}{2}=45 \text { ways. }
\)
Example 29: A committee of 5 members must be chosen from a track club. The club has 15 sprinters, 9 jumpers, and 7 long-distance runners. The committee must have exactly 1 jumper and 1 long-distance runner. How many ways may the committee be chosen?
Solution:
Step 1: Given, we have three different sets in this case: 15 sprinters, 9 jumpers, and 7 long-distance runners.
Step 2: We want to create a set of size 5 from 3 smaller sets: 3 sprinters, 1 jumper, and 1 long-distance runner.
Step 3: We will be multiplying 3 different combinations together:
\(
\begin{aligned}
& \frac{15 !}{3 !(15-3) !} \times \frac{9}{1 !(9-1) !} \times \frac{7}{1 !(7-1) !}=\frac{15 \times 14 \times 13 \times 12 !}{3 ! \times 12 !} \times \frac{7 \times 6 !}{6 !} \times \frac{7 \times 6 !}{6 !} \\
& =\frac{15 \times 14 \times 13}{3 \times 2} \times 9 \times 7 \\
& \mathbf{= 2 8}, \mathbf{6 6 5} \text { committees. } \\
&
\end{aligned}
\)
Example 30: How many different car license plates can be made with one letter followed by one number?
Solution:
There are 26 choices of letters. Each of these 26 letters can be followed by one of 10 different digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Therefore there are 26 x1 0 = 260 different plates which can be made.
Example 31: How many different car license plates can be made with two letters followed by three numbers?
Solution:
There are 26 choices for the first letter. For each of these letters, there are 26 choices for the second letter. There are therefore \(26 \times 26=676\) possible pairs of letters. (Note that a repeat letter, such as DD, is allowed and so we do not use \({ }^{26} \mathrm{P}_2\).)
You must now consider the three numbers. There are 10 possibilities for the first digit, 10 possibilities for the second, and 10 possibilities for the third. This means that there are \(10 \times 10 \times 10=1000\) different numbers. (Note that this is simply saying that there are 1000 numbers between and including 000 and 999.)
Combining these results, it follows that there are \(676 \times 1000=676,000\) different license plates possible.
This is when the elements of a set can be repeated, to clarify this type, here is an example: A person goes to a candy shop, where there are 10 different flavors of candy, but this person is only going to take 4, one for each one of his children, this is an example of combination with repetition because although there are 10 different flavors, anything disallows this person to pick the same flavor twice, trice or even four times.
Formula with repetition
\(Example 32: A person is going to a candy shop where there are 8 types of flavors, if this person is only going to buy 3, define every combination possible.
Solution:
In this case \(n=8\) and \(r=3\)
\(Example 33: A sportsman goes to the store to buy 4 pairs of shoes, if at the store there are a lot of shoes in 5 available colors, how many combinations of colors can this man buy?
Solution:
In this case \(n=5\) and \(r=4\)
\(You cannot copy content of this page