Question 1)
By applying the formula for combinations, calculate the following

C (8, 3)

The formula for combinations is
Thus,
Simplifying it, we obtain C (8, 3) = 56

C (12, 6)

The formula for combinations is
Thus,
Simplifying it, we obtain C (12, 6) = 924

C (9, 3)

The formula for combinations is
Thus, Thus,
Simplifying it, we obtain 84

Question 2) Find the number of combinations of

Five objects taken two at a time

Total objects, n = 5
Objects in each group, r = 2
Applying the formuApplying the formula for combinations,
Simplifying, total combinations = 10

Ten objects taken three at a time

Total objects, n = 10
Objects in each group, r = 3
Applying the formuApplying the formula for combinations,
Simplifying, total combinations = 120

Fifteen objects taken four at a time

Total objects, n = 15
Objects in each group, r = 4
Applying the formuApplying the formula for combinations,
Simplifying, total combinations = 1365

Question 3) In a class of fifty students, ten students decide to go on a trip. Solve the following:

Calculate the number of different groups of students that can go on the trip

The total number of students is 50, out of which, you have to make different selections
or groups of ten students. How many such groups can be made? This question can be
answered by the formula of combinations as follows:
Total number of students, n = 50
Number of students in each group, r = 10
Applying the formula Applying the formula of combinations,
Simplifying, total combinations = 10,272,278,170 Therefore 10,272,278,170 different
groups of students can be formed that can go on the trip.

Find out the number of different groups of students that will possibly not go on the trip

For each selection of 10 students that will go on the trip, 40 students will not
be going on the trip because there are a total of 50 students.
Total number of students, n = 50
Number of students in each group that will not go on the trip, r = 40
Applying the formula for combinations, Applying the formula for combinations,
Simplifying, total combinations = 10,272,278,1700,272,278,170 different groups of
students can be formed that will not go on the trip.

If two head students are chosen to go on the trip necessarily, and eight other students
are to be chosen randomly, what are the number of different groups of students that
can go on the trip?

Since two students are already present in each group of 10 students, therefore we
have to find the number of different selections of 8 students from the remaining
48 students.
Thus, total number of students, n = 48
Number of students in each group, r = 8
Applying the formula for combinations, Applying the formula for combinations,
Simplifying, total combinations = 377,348,9948,994 different groups of students
can be chosen in which two head students will always be included.

Question 4) Solve the following word
problems on combinations:

From a class of twenty students, half are to be selected for a drawing competition.
If three students agree that either they will all go in the competition together,
or they will not go, find the number of different groups (combinations) of students
that can be selected for the competition.

This problem takes into consideration two different types of selections:

The three students will go in the competition

In this case, since three of the twenty students have already been taken, number
of students left, n = 17
Number of students to be chosen for each group, r = 7
Applying the formula for combinations,
Simplifying, total combinations = 19,448 different groups of students can be possibly
chosen for the competition.

The three students will not go in the competition

In this case, the three students will not go in the competition, and hence, total
number of students left to choose from, n = 17

On the other hand, all ten students are to be chosen, so number of students in each
group/selection, r = 10

Applying the formula for combinations,

Simplifying, total combinations = 19,448 Therefore 19,448 different groups of students
can be chosen to to in the competition.

Determine the number of different selections of four red cards possible from a pack
of 52 cards.

In a pack of 52 cards, there are 26 red and 26 black cards. Hence,
Total number of cards to choose from, n = 26
Number of cards in each group, r = 4

Applying the formula for combinations,

Simplifying, total combinations = 14,950. Therefore 14,950 can be selected from
a pack of 52 cards.

If, from a pack of 52 cards, only four are to be selected such that each of the
four cards belongs to a different suite, find the number of different selections
that can be made for four such cards

There are four suites of thirteen cards each in a pack of 52 cards. From each suite,
one card can be selected in thirteen different ways, since there are thirteen choices
for one card. Thus for four cards of each different suite, the total number of different
selections possible are 13 x 13 x 13 x 13, which equals 28,561.

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