Five girls- Fiorenza, Gladys, Helene, Jocelyn, and Kaitlin - and four boys- Abe, Bruce, Clive, and Doug ride to school each day in three seperate buses.
Abe and Fiorenza always ride together.
Gladys and Helen always ride together.
Jocelyn and Kaitlin never ride together.
Doug always rides in the bus with the fewest children.
Boys cannot outnumber girls in any bus.
The maximum number of children in any bus is four.
The bus in which Doug rides can hold how many children?
Bruce can ride with each of the following except
a.Abe
b.Clive
c.Doug
f.Fiorenza
e.Helene
To make it easier to read, I used initials instead of names:
Five girls- F, G, H, J, and K - and four boys- A, B, C, and D
ride to school each day in three separate buses.
1. A and F always ride together.
2. G and H always ride together.
3. J and K never ride together.
4. D always rides in the bus with the fewest children.
5. Boys cannot outnumber girls in any bus.
6. The maximum number of children in any bus is four.
The bus in which D rides can hold how many children?
Let's see if we can all the ways to seat all 9 children in the buses
X, Y, and Z following those rules.
Make this table:
Bus X Bus Y Bus Z
Girls
Boys
I'll start by putting A,F,G,H on Bus X, which fits 1 and 2
Bus X Bus Y Bus Z
Girls FGH
Boys A
Now because of 3, we have to put J and K on different buses.
And because of 6, neither can go on Bus X
So let's put J in Bus Y and K in Bus Z.
Bus X Bus Y Bus Z
Girls FGH J K
Boys A
That only leaves boys B, C, D.
I can only put one of those three boys on Bus Y because of 5.
So I'll put B on Bus Y and the other two boys on Bus Z. But
that won't do because there'd be no place to put D, since
D has to be on the bus with the fewest children. D can't go
on Bus Y and leave K alone for she would then be on the bus
with the least number of children. And it's the same thing
if you put him in Bus Z. So it's not possible for J and K
to be on buses Y and Z.
So it is impossible to seat all 9 when F,G,H,A are all
on the same bus, So let's start over, putting F,G on one bus
and H,A on another.
Bus X Bus Y Bus Z
Girls FG H
Boys A
Now we seat J and K. By 3, there are 3 ways to seat them. This is
the first way:
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A
That leaves B, C, D, Then because of 4, D must go in Bus Z
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A D
Now we have three ways to place B and C. This way:
Girls FGJ HK
Boys B AC D
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I underlined it with stars to show it is a solution.
This way
Girls FGJ HK
Boys B A CD
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And this way:
Girls FGJ HK
Boys AB CD
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Now go back to where we putting F,G on one bus
and H,A on another.
Bus X Bus Y Bus Z
Girls FG H
Boys A
Now we seat J and K. By 3, there are 3 ways to seat them. This is
the second way:
Bus X Bus Y Bus Z
Girls FGJ H K
Boys A
This leaves boys B, C, D. We can't put any of these boys on Bus Y or
they'd outnumber girl H. So if we put D on Bus Z then D wouldn't be
on the bus with the fewest children, and we can't put D on Buses X or Y.
So we can't put J and K on Buses X and Z. So this is ruled out. So
that leaves only this way to place J and K on different buses:
Bus X Bus Y Bus Z
Girls FG HJ K
Boys A
This leaves boys B, C, D. D obviously has to go in Bus Z or D wouldn't be
on the bus with the fewest children.
So we have
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys A
Now one of B,C must go on Bus X for it to have more children than
the bus D is on.
So we have
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys B A
Now C can either go on Bus X or Y. Either this
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys BC A
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or
Bus X Bus Y Bus Z
Girls FG HJ KD
Boys B AC
***********************************************************
That leaves B, C, D, Then because of 4, D must go in Bus Z
Bus X Bus Y Bus Z
Girls FGJ HK
Boys A D
Now we have three ways to place B and C. This way:
Girls FGJ HK
Boys B AC D
********************************************************
I underlined it with stars to show it is a solution.
This way
Girls FGJ HK
Boys B A CD
********************************************************
And this way:
Girls FGJ HK
Boys AB CD
******************************************************
Now we've found all possible solutions to the seating
arrangements on the busses.
The first question is:
The bus in which Doug rides can hold how many children?
From all the solutions starred, D is on bus Z either
by himself, with C, or with D. So at most 2 children
can be on the bus D rides on.
Bruce can ride with each of the following except ___
Look at all the solutions and you'll see there is no
solution that has B on the same bus with D.
Edwin