20 different comic books will be distributed to five kids. (a) how many ways are there to distribute the comic books if there are no restrictions on how many go to each kid (other than the fact that all 20 will be given out)? (b) how many ways are there to distribute the comic books if they are divided evenly so that 4 go to each kid?

a) 21684

b) 15504

Step-by-step explanation:

a) Distribute the comic books as follows

only to 1 kid          one way   but we have five kids then  5

to two kids                         C₂₀,₂  =  20! /2! ( 20-2)!    =  20*19 /2  =   190

To three   C₂₀,₃   =  20! /3! ( 20 - 3)!       =  20*19*18 /6               =  1140

To four     C₂₀,₄    =   20 ! / 4! ( 20 - 4 )!   = 20*19*18*17 /4*3*2   =  4845

To five     C₂₀,₅    =    20!  /5! (20 - 5 )!    = 20*19*18*17*16/5*4*3*2*1

C₂₀,₅    =  15504

Then total ways of distribution are:

5 + 190 + 1140 + 4845 + 15504   =  21684

b)   C₂₀,₅  we know from a    that is equal to  15504

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3