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?

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 = 14.13.12! = 14.13 = 91

           12! 2!            2    

C₁₆,₂ = 16.15.14! = 16.15 = 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 = 14.13.12.11.10! = 14.13.12.11 = 1001

                  10! 4!               4.3.2.1    

C₁₆,₄ = 16.15.14.13.12! = 16.15.14.13 = 1820

                  12! 4!               4.3.2.1    

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

x=-7

step-by-step explanation:

x+3=-10

-10+3=-7

x=-7