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\(
{ }^8 C_3=?
\)
\(
{ }^8 C_3=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56
\)
In how many ways can a cricket eleven be chosen out of 14 players?
\(
\text { Required number of ways }={ }^{14} C_{11}={ }^{14} C_{(14-11)}={ }^{14} C_3=\frac{14 \times 13 \times 12}{3 \times 2 \times 1}=364
\)
In how many ways, a committee of 6 members be selected from 7 men and 5 ladies, consisting of 4 men and 2 ladies?
We have to select (4 men out of 7) and (2 ladies out of 5).
\(
\therefore \text { Required number of ways }={ }^7 C_4 \times{ }^5 C_2={ }^7 C_3 \times{ }^5 C_2=\left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)=350
\)
On the plane, there are 6 different points (no three of them are lying on the same line). How many segments do you get by joining all the points?
\(
\begin{aligned}
& C(6.2)=\frac{6 !}{4 ! 2 !} \\
& C(6.2)=\frac{6.5}{2} \\
& C(6.2)=15
\end{aligned}
\)
On a circle, there are 9 points selected. How many triangles with edges in these points exist?
\(
\begin{aligned}
& C(3,9)=\frac{9 !}{6 ! 3 !} \\
& C(3,9)=\frac{9.8 .7 .6 !}{6 ! 6} \\
& C(3,9)=\frac{9.8 .7}{6} \\
& C(3,9)=84
\end{aligned}
\)
How many diagonals have a 10-gon?
\(
\begin{aligned}
N=\left(\begin{array}{c}
10 \\
2
\end{array}\right)-10=45-10=35 \\
N=35
\end{aligned}
\)
Convex 10-gon has 35 diagonals.
Formula to calculate is below:
\(
N=\frac{n(n-3)}{2}
\)
In how many ways you can choose 8 of 32 playing cards not considering their order?
\(
\begin{aligned}
& C(8,32)=\left(\begin{array}{c}
32 \\
8
\end{array}\right) \\
& C(8,32)=\frac{32 !}{24 ! 8 !} \\
& C(8,32)=\frac{32 \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\
& C(8,32)=10518300
\end{aligned}
\)
The playing cards can be chosen in 10518300 ways.
A teacher has prepared 20 arithmetics tasks and 30 geometry tasks. For a test he‘d like to use 3 arithmetics and 2 geometry tasks. How many ways are there to build the test?
\(
\begin{aligned}
& N=\left(\begin{array}{c}
20 \\
3
\end{array}\right)\left(\begin{array}{c}
30 \\
2
\end{array}\right) \\
& N=\frac{20 !}{17 ! \cdot 3 !} \cdot \frac{30 !}{28 ! \cdot 2 !} \\
& N=1140 \times 435 \\
& N=495900
\end{aligned}
\)
On a graduation party, the graduates pinged their glasses. There were 253 pings. How many graduates came to the party?
\(
\begin{aligned}
& C(2, x)=253 \\
& \left(\begin{array}{l}
x \\
2
\end{array}\right)=253 \\
& \frac{x !}{(x-2) ! 2 !}=253 \\
& \frac{x(x-1)}{2}=253 / 2 \\
& x^2-x=506 \\
& x^2-x-506=0 \\
& (x-23)(x+22)=0 \\
& x_1=23 \\
& x_2=-22 \text { not possible } \\
& K=\{23\}
\end{aligned}
\)
There were 23 graduates at the party.
In the confectioners, 5 different icecreams are sold. A father would like to buy 15 caps of ice cream for his family. In how many ways can he buy the ice cream?
\(
\begin{aligned}
& C(15,5)=\left(\begin{array}{c}
5+15-1 \\
15
\end{array}\right) \\
& C(15,5)=\left(\begin{array}{l}
19 \\
15
\end{array}\right) \\
& C(15,5)=\frac{19 !}{4 ! \cdot 15 !} \\
& C(15,5)=\frac{19 \cdot 18 \cdot 17 \cdot 16}{4 \cdot 3 \cdot 2 \cdot 1} \\
& C(15,5)=3876
\end{aligned}
\)
Over the weekend, your family is going on vacation, and your mom is letting you bring your favorite video game console as well as five of your games. How many ways can you choose the five games if you have 12 games in all?
\(
{ }^{12} C_5=(12) ! /(5 ! \times(12-5) !)=(12) ! /(5 ! \times(7) !)=792
\)
How many different groups of 10 students can a teacher select from her classroom of 15 students?
\(
{ }^{15} C_{10}=\frac{15 !}{(15-10) ! 10 !}=3003
\)
\(
\text { If }{ }^n C_{12}={ }^n C_9 \text { find }{ }^{21} C_n
\)
\(
\begin{aligned}
& { }^n C_{12}=n ! /(n-12) ! 12 ! \dots(1) \\
& { }^n C_9=n ! /(n-9) ! 9 ! \quad \cdots(2) \\
& n ! /(n-12) ! 12 !=n ! /(n-9) ! 9 ! \\
& (n-9) ! 9 !=(n-12) ! 12 ! \\
& (n-9)(n-10)(n-11)(n-12) ! 9 !=(n-12) ! 12 \cdot 11 \cdot 10 \cdot 9 ! \\
& (n-9)(n-10)(n-11)=12 \cdot 11 \cdot 10 \\
& n-9=12 \\
& n=12+9 \\
& n=21
\end{aligned}
\)
\(
{ }^{21} c_n={ }^{21} c_{21}=1
\)
Hence the answer is 1.
\(
\text { If }{ }^{15} C_{2 r-1}={ }^{15} C_{2 r+4} \text {, find } r \text {. }
\)
\(
\begin{gathered}
\text { If }{ }^n C_x={ }^n C_y \Rightarrow x=y \text { (or) } x+y=n \\
2 r-1+2 r+4=15 \\
4 r+3=15 \\
4 r=12
\end{gathered}
\)
Divide by 4 on both sides.
\(
r=12 / 4 \Rightarrow 3
\)
Hence the value of \(r\) is 3 .
In how many ways any 3 boys of a class of 20 boys are arranged in order of their increasing height? Assume that all the boys in the class had different heights.
\(
\text { Number of ways in which any } 3 \text { boys can be selected }={ }^{20} C_3=1140 \text {. }
\)
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be
chosen in \({ }^7 \mathrm{C}_3\) ways.
\(
{ }^7 C_3=\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 !}{3 \cdot 2 \cdot 1 \cdot 4 !}=35
\)
How many committees of 5 members can be chosen from a group of 8 persons when each committee must include 2 particular persons?
We have to select 6 persons out of 3 members and this can be done in \({ }^6 \mathrm{C}_3\) ways. Thus the numbers of committees are
\(
={ }^{8-2} C_{5-2}=\frac{6 !}{3 !(6-3) !}=\frac{6 \cdot 5 \cdot 4 \cdot 3 !}{3 \cdot 2 \cdot 1 \cdot 3 !}=5 \times 4=20
\)
A team of 5 basketball players including a captain needs to be formed out of 13 basketball players Find the number of ways in which this job can be done.
A team of 5 basketball players can be selected from 13 players \(={ }^{13} C_5\). Now there can be 5 different captains in each team. Hence the total number of ways \(={ }^{13} \mathrm{C}_5 \times 5\).
In how many ways 8 boys can be distributed in two teams of 4 each to play a game of cricket?
Four boys can be selected from 8 in \({ }^8 \mathrm{C}_4\) ways. The other 4 boys will form another team. Since, in this each group of 4 is considered twice, eg. Suppose there are 8 boys 1 to 8 . One time a group will be formed from 1-4 and other time a group will be formed from 5 8. Both these groups will form same two pairs of teams. Hence, the total number of ways \(=1 / 2 \times{ }^8 C_4\) ways.
James, Taylor, Steve, and Chris are four soccer players and we want to build a team consisting of three players. In how many ways this team can be formed?
We take initials of each name as \(J, T, S\), and \(C\)
We know that order of arrangement is not important in combination related problems
So team can be built as \(J T S, J T C, J S C\) and \(T S C\). Here \(J T S=S T J=T J S\) so we do not include such arrangements while calculating combination numerical.
So,
\(
\begin{aligned}
& C(4,3)=\frac{4 !}{3 !(4-3) !)} \\
& C(4,3)=\frac{4 \times 3 !}{3 !} \\
& C(4,3)=4
\end{aligned}
\)
A Science club in a school consists of 18 girls and 10 boys. In how many ways a team of 5 girls and 3 boys can be formed?
The number of combinations of 7 girls out of 18 can be written as \(C(18,7)\)
Similarly the number of combinations of 3 boys out of 10 can be written as \(C(10,3)\)
We want to find in how many ways they can for a team. We can do it by simply multiplying the both combination
\(
\begin{aligned}
C(18,7) \times C(10,3)= & {\left[\frac{18 !}{7 !(18-5) !}\right]\left[\frac{10 !}{3 !(10-3) !}\right] } \\
& =\left[\frac{18 !}{7 !(13) !}\right]\left[\frac{10 !}{3 !(7) !}\right] \\
& =(204)(120) \\
& =244804.
\end{aligned}
\)
A safe is to be locked with a four-digit password. The digits must contain numbers from 0−7. How many different password options you will have if no order arrangement is followed? ( Combination Problem)
In this first case, the order does not matter and we have to choose four digits out of 8.
\(
C(8,4)=\frac{8 !}{4 !(8-4) !)}=70
\)
A safe is to be locked with a four-digit password. The digits must contain numbers from 0−7. How many different password options you will have if the required arrangement is to be followed? ( Permutation Problem)
Here the order of selection matters and hence we use the formula for permutation, i.e.,
\(
P(8,4)=\frac{8 !}{(8-4) !)}=1680
\)
A soccer team consisting of 11 players is to be formed from a pool of 15 players. In how many ways the team can be selected if the Captain of the team is to be included in every team?
If Captain is to be included in every team selection then we have to select 10 players from a pool of 14 players:
\(
C(14,10)=\frac{14 !}{10 !(14-10) !)}=1001
\)
In how many ways a committee of 6 boys and 4 girls can be chosen from 12 boys and 6 girls? In how many ways the committee will be formed if a particular boy and girl has to be included in each committee?
A committee consisting of 6 boys and 4 girls can be calculated as
\(
\begin{aligned}
& C(12,6) \times C(6,4)=\left[\frac{12 !}{6 !(12-6) !}\right]\left[\frac{6 !}{4 !(6-4) !}\right] \\
& =\left[\frac{12 !}{6 ! \times 6 !}\right]\left[\frac{6 !}{4 !(2) !}\right] \\
& =(924)(15) \\
& =13860 \text { ways }
\end{aligned}
\)
If a particular boy and a girl is included in each committee then we can calculate as
\(
\begin{aligned}
& C(11,5) \times C(5,3)=\left[\frac{11 !}{5 !(11-5) !}\right]\left[\frac{5 !}{3 !(5-4) !}\right] \\
& =\left[\frac{11 !}{5 ! 6 !}\right]\left[\frac{5 !}{3 !(2) !}\right]
\end{aligned}
\)
\(
\begin{aligned}
& =(462)(10) \\
& =4620 \text { ways }
\end{aligned}
\)
\(
\text { If } C(n, 12)=C(n, 16) \text { calculate the value of } n \text { ? }
\)
By the relation of complimentary combintions
\(
C(n, k)=C(n, n-k)
\)
So,
\(
C(n, 12)=C(n, n-12)
\)
And we have been given that \(C(n, 12)=C(n, 16)\)
So,
\(
C(n, n-12)=C(n, 16)
\)
So we have same indices and base and we can calculate the value of \(n\) as
\(
\begin{aligned}
& n-12=16 \\
& n=16+12 \\
& n=28
\end{aligned}
\)
How many triangles can be formed by joining the vertices of an Octagon?
We know that the triangle has 3 vertices so for given amount of vertices we have to select 3 to form a triangle.
\(
C(8,3)=\frac{8 !}{3 !(8-3) !)}=56
\)
How many Diagonals can be formed by joining the vertices of a Decagon?
For Decagon we have 10 sides.
\(
\text { Total number of line segments: } \quad C(10,2)=\frac{10 !}{2 !(10-2) !)}=45
\)
\(
\text { Total number of diagonals }=45-10=35
\)
Find the total number of subsets of a set with 4 elements.
\(
\left[\text { Hint: }{ }^n C_0+{ }^n C_1+{ }^n C_2+\cdots+{ }^n C_n=2^n\right]
\)
\(
{ }^n C_0+{ }^n C_1+{ }^n C_2+\cdots+{ }^n C_n=2^n
\)
here \(n=4\)
\(
\begin{aligned}
={ }^4 C_0+{ }^4 c_1 & +{ }^4 C_2+{ }^4 C_3+{ }^4 C_4 \\
& =2^4 \\
& =16
\end{aligned}
\)
How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?
Out of 10 members, we have to select a chair person. So we have 10 options to select a chair person. 9 options to select a secretary.
After selecting a chairperson and secretary, we have 8 members. Out of 8, we have to select 4 persons.
Hence the answer is \((10 \cdot 9)^8 \mathrm{C}_4\)
How many different selections of 5 books can be made from 12 different books if two particular books are always selected?
Since two particular books are always selected, we may select remaining 3 books out of 10 books.
\(
\begin{aligned}
{ }^{10} C_3=10 ! /(7 ! 3 !) & =(10 \cdot 9 \cdot 8) /(3 \cdot 2) \\
& =120
\end{aligned}
\)
Hence the answer is 120.
How many different selections of 5 books can be made from 12 different books if two particular books are never selected?
Since two particular books are never selected, we may select 5 books out of 10 books.
\(
\begin{gathered}
{ }^{10} C_5=10 ! /(5 ! 5 !)=(10 \cdot 9 \cdot 8 \cdot 7 \cdot 6) /(5 \cdot 4 \cdot 3 \cdot 2) \\
=252
\end{gathered}
\)
Mark has 5 pants and 7 shirts in his closet. He wants to wear a different pant/shirt combination each day without buying new clothes for as long as he can. How many weeks can he do this for?
Mark has 35 unique combinations to choose from, so can he co go 5 weeks with this set of clothes
Number of pants in Mark’s closet \(=5\)
Number of shirts in Mark’s closet \(=7\)
Combinations are ways to choose elements from a collection in mathematics where the order of the selection is irrelevant. Let’s say we have a trio of numbers: \(\mathrm{P}, \mathrm{Q}\), and \(\mathrm{R}\). Then, combination determines how many ways we can choose two numbers from each group.
He can choose amongst the pants in 5 ways, similarly, he can choose among the shirts in 7 ways
Total number of combinations \(=5 \times 7=35\) unique combinations
So, if he wears one combination each day he can last 35 days or 5 weeks, without buying new clothes.
Twenty students enter a contest at school. The contest offers a first, second, and third prize. How many different combinations of 1st, 2nd, and 3rd place winners can there be?
This is a permutation problem because we are looking for the number of groups of winners. Consider the three positions, and how many choices there are for each position: There are 20 choices for 1st place, 19 for 2nd place, and 18 for 3rd place.
20, 19, 18
Multiply to get 6840.
A baker has four different types of frosting, three different kinds of sprinkles, and 8 different cookie cutters. How many different cookie combinations can the baker create if each cookie has one type of frosting and one type of sprinkle?
Since this a combination problem and we want to know how many different ways the cookies can be created we can solve this using the Fundamental counting principle. 4 x 3 x 8 = 96
Multiplying each of the possible choices together.
At an ice cream store, there are 5 flavors of ice cream: strawberry, vanilla, chocolate, mint, and banana. How many different 3-flavor ice cream cones can be made?
There are \(5 \times 4 \times 3\) ways to arrange 5 flavors in 3 ways. However, in this case, the order of the flavors does not matter (e.g., a cone with strawberry, mint, and banana is the same as a cone with mint, banana, and strawberry). So we have to divide \(5 \times 4 \times 3\) by the number of ways we can arrange 3 different things which is \(3 \times 2 \times 1\). So \((5 \times 4 \times 3) /(3 \times 2 \times 1)\) is 10.
One can also use the combination formula for this problem: \({ }^n C_r=n ! /(n-r) ! r !\)
Therefore: \({ }^5 \mathrm{C}_3=5 ! / 3 ! 2 !\)
\(=10\)
At a deli, you can choose from either Italian bread, whole wheat bread, or sourdough bread. You can choose turkey or roast beef as your meat and provolone or mozzarella as your cheese. If you have to choose a bread, a meat, and a cheese, how many possible sandwich combinations can you have?
You have 3 possible types of bread, 2 possible types of meat, and 2 possible types of cheese. Multiplying them out you get 3 x 2 x 2, giving you 12 possible combinations.
Shannon decided to go to nearby cafe for lunch. She can have a sandwich made on either wheat or white bread. The café offers cheddar, Swiss, and American for cheese choices. For meat, Shannon can choose ham, turkey, bologna, roast beef, or salami. How many cheese and meat sandwich options does Shannon have to choose from?
2 bread choices x 3 cheese choices x 5 meat choices = 30 sandwich choices
An ice cream parlor serves 36 ice cream flavors. You can order any flavor in a small, medium or large and can choose between a waffle cone and a cup. How many possible combinations could you possibly order?
36 possible flavors x 3 possible sizes x 2 possible cones = 216 possible combinations.
Tifa’s restaurant offers a special in which a patron may choose from a selection of 4 appetizers, 7 entrees, and 3 desserts. How many combinations of 1 appetizer, 1 entree, and 1 dessert are possible?
Just multiply to find the possible combinations:
\(
4 \times 7 \times 3=84
\)
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