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damaskus [11]
3 years ago
9

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?
Mathematics
1 answer:
Dmitrij [34]3 years ago
7 0

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

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

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

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?

16.28.27.26.15 = 4,717,440

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

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = <u>14.13.12!</u> = <u>14.13 </u>= 91

           12! 2!            2    

C₁₆,₂ = <u>16.15.14!</u> = <u>16.15 </u>= 120

           14! 2!            2    

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

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

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = <u>14.13.12.11.10!</u> = <u>14.13.12.11 </u>= 1001

                  10! 4!               4.3.2.1    

C₁₆,₄ = <u>16.15.14.13.12!</u> = <u>16.15.14.13 </u>= 1820

                  12! 4!               4.3.2.1    

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

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