There are 14 juniors and 16 seniors in a chess club. a) from the 30 members, how many ways are there to arrange 5 members of the club in a line? b) how many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line? 0 c) if the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there? d) if the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

No. of juniors = 14

No. of seniors = 16

Total students = 30

A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

Since we are asked about arrangement so we will use permutation

Formula :

n = 30

r = 5

So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?

B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

Out of 16 seniors 2 will be selected

So, 3 places are vacant

Remaining students = 30-2 = 28

So, out of 28 students 3 students will be selected

No. of ways =

No. of ways =

                   =

There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.

C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Since we are not asked about arrangement so we will use combination

Out of 16 seniors 2 will be selected

Out of 14 juniors 2 will be selected

Formula :

So, No. of possible groupings =

                                                  =

                                                  =

If the club sends 2 juniors and 2 seniors to the tournament, there are 10920 possible groupings

D) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Out of 16 seniors 4 will be selected

or

Out of 14 juniors 4 will be selected

So, No. of possible groupings =

                                                  =

                                                  =

So,If the club sends either 4 juniors or 4 seniors, there are 2821 possible groupings .

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value