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junior college 2 | H2 Maths
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I got 40 but not sure if I’m correct.
First carriage can be arranged in 2 ways
After taking out 3 carriages from remaining 9, left with 6 carriages, 7 slots to put 3 carriages into so 7C3=35, then 3 ways to arrange among the 3 carriages so 35+2+3=40
Date Posted: 3 years ago
Views: 262
Carriages :
A B C D E F G H I J
Let the first carriage be A.
Slotting method :
So from carriage B to carriage J, we need 3 carriages to have the dispenser.
We 'slot' them between the other 6 carriages.
There are 7 possible slots so 7C3 = 35
Note that there is no need to permutate the three carriages as they are fixed already, since each carriage is unique and distinguishable.
A B C D E F G H I J
Let the first carriage be A.
Slotting method :
So from carriage B to carriage J, we need 3 carriages to have the dispenser.
We 'slot' them between the other 6 carriages.
There are 7 possible slots so 7C3 = 35
Note that there is no need to permutate the three carriages as they are fixed already, since each carriage is unique and distinguishable.
Cases method (A will not be shown for clarity)
Case 1 : The dispenser-carrying carriages are only one carriage apart.
Eg. [B] C [D] E [F] G H I J
(Brackets indicates the dispenser)
Including the ones in between them, number of carriages = 5
Since there are 9 carriages, we can shift the dispensers rightwards 4 times to the end.
B [C] D [E] F [G] H I J
B C [D] E [F] G [H] I J
B C D [E] F [G] H [I] J
B C D E [F] G [H] I [J]
number of ways here = 5
Case 2 : They are two carriages apart
Eg. [B] C D [E] F G [H] I J
Including the ones in between them, number of carriages = 7
Since there are 9 carriages, we can shift the dispensers rightwards 2 times to the end.
number of ways = 3
Case 3 : they are 3 carriages apart
Eg. [B] C D E [F] G H I [J]
Number of ways = 1
Case 1 : The dispenser-carrying carriages are only one carriage apart.
Eg. [B] C [D] E [F] G H I J
(Brackets indicates the dispenser)
Including the ones in between them, number of carriages = 5
Since there are 9 carriages, we can shift the dispensers rightwards 4 times to the end.
B [C] D [E] F [G] H I J
B C [D] E [F] G [H] I J
B C D [E] F [G] H [I] J
B C D E [F] G [H] I [J]
number of ways here = 5
Case 2 : They are two carriages apart
Eg. [B] C D [E] F G [H] I J
Including the ones in between them, number of carriages = 7
Since there are 9 carriages, we can shift the dispensers rightwards 2 times to the end.
number of ways = 3
Case 3 : they are 3 carriages apart
Eg. [B] C D E [F] G H I [J]
Number of ways = 1
Case 4 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 2 carriages apart
Eg. [B] C [D] E F [G] H I J
Number of ways = 4
Case 5 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 3 carriages apart
Eg. Eg. [B] C [D] E F G [H] I J
Number of ways = 3
Case 6 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 4 carriages apart
Eg. Eg. [B] C [D] E F G H [I] J
Number of ways = 2
Case 7 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 5 carriages apart
Eg. [B] C [D] E F G H I [J]
Number of ways = 1
Eg. [B] C [D] E F [G] H I J
Number of ways = 4
Case 5 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 3 carriages apart
Eg. Eg. [B] C [D] E F G [H] I J
Number of ways = 3
Case 6 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 4 carriages apart
Eg. Eg. [B] C [D] E F G H [I] J
Number of ways = 2
Case 7 : first two dispenser-carriages 1 carriage apart, 2nd and 3rd are 5 carriages apart
Eg. [B] C [D] E F G H I [J]
Number of ways = 1
Case 8 : first two dispenser-carriages 2 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D [E] F [G] H I J
Number of ways = 4
Case 9 : first two dispenser-carriages 2 carriages apart, 2nd and 3rd are 3 carriages apart
Eg. [B] C D [E] F G H [I] J
Number of ways = 2
Case 10 : first two dispenser-carriages 2 carriages apart, 2nd and 3rd are 4 carriages apart
Eg. [B] C D [E] F G H I [J]
Number of ways = 1
Eg. [B] C D [E] F [G] H I J
Number of ways = 4
Case 9 : first two dispenser-carriages 2 carriages apart, 2nd and 3rd are 3 carriages apart
Eg. [B] C D [E] F G H [I] J
Number of ways = 2
Case 10 : first two dispenser-carriages 2 carriages apart, 2nd and 3rd are 4 carriages apart
Eg. [B] C D [E] F G H I [J]
Number of ways = 1
Case 11 : first two dispenser-carriages 3 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D E [F] G [H] I J
Number of ways = 3
Case 12 : first two dispenser-carriages 3 carriages apart, 2nd and 3rd are 2 carriages apart
Eg. [B] C D E [F] G H [I] J
Number of ways = 2
Eg. [B] C D E [F] G [H] I J
Number of ways = 3
Case 12 : first two dispenser-carriages 3 carriages apart, 2nd and 3rd are 2 carriages apart
Eg. [B] C D E [F] G H [I] J
Number of ways = 2
Case 13.
two dispenser-carriages 4 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D E F [G] H [I] J
Number of ways = 2
Case 14 : first two dispenser-carriages 4 carriages apart, 2nd and 3rd are 2 carriages apart
Eg. [B] C D E F [G] H I [J]
Number of ways = 1
Case 15 : first two dispenser-carriages 5 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D E F G [H] I [J]
Number of ways = 1
two dispenser-carriages 4 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D E F [G] H [I] J
Number of ways = 2
Case 14 : first two dispenser-carriages 4 carriages apart, 2nd and 3rd are 2 carriages apart
Eg. [B] C D E F [G] H I [J]
Number of ways = 1
Case 15 : first two dispenser-carriages 5 carriages apart, 2nd and 3rd are 1 carriage apart
Eg. [B] C D E F G [H] I [J]
Number of ways = 1
Total ways :
5 + 3 + 1 + 4 + 3 + 2 + 1 + 4 + 2 + 1 + 3 + 2 + 2 + 1 + 1
= 35
5 + 3 + 1 + 4 + 3 + 2 + 1 + 4 + 2 + 1 + 3 + 2 + 2 + 1 + 1
= 35
We notice a pattern :
1 apart, 5 apart and 5 apart, 1 apart are mirrored. They have the same number of ways.
2 apart , 3 apart and 3 apart, 2 apart are also mirrored.
Same for 1,4 and 4,1 , 2,4 and 4,2, etc.
The only ones that don't have mirrors are 1,1 2,2 and 3,3
Number of ways can rewritten as :
(1 + 2) + (1 + 2 + 3 + 4) + (4 + 3 + 2 + 1) + (2 + 1) + 1 + 3 + 5
= 3 + 10 + 10 + 3 + 1 + 3 + 5
= 26 + 9
= 35
1 apart, 5 apart and 5 apart, 1 apart are mirrored. They have the same number of ways.
2 apart , 3 apart and 3 apart, 2 apart are also mirrored.
Same for 1,4 and 4,1 , 2,4 and 4,2, etc.
The only ones that don't have mirrors are 1,1 2,2 and 3,3
Number of ways can rewritten as :
(1 + 2) + (1 + 2 + 3 + 4) + (4 + 3 + 2 + 1) + (2 + 1) + 1 + 3 + 5
= 3 + 10 + 10 + 3 + 1 + 3 + 5
= 26 + 9
= 35
Or if we list them in order
1,1 1,2 1,3 1,4 1,5
2,1 2,2 2,3 2,4
3,1 3,2 3,3
4,1, 4,2
5,1
We would get :
5 + 4 + 3 + 2 + 1 →15 ways
4 + 3 + 2 + 1 →10 ways
3 + 2 + 1 →6 ways
2 + 1 →3 ways
1 →1 ways
Total = 35
This is known as summation of summations.
1,1 1,2 1,3 1,4 1,5
2,1 2,2 2,3 2,4
3,1 3,2 3,3
4,1, 4,2
5,1
We would get :
5 + 4 + 3 + 2 + 1 →15 ways
4 + 3 + 2 + 1 →10 ways
3 + 2 + 1 →6 ways
2 + 1 →3 ways
1 →1 ways
Total = 35
This is known as summation of summations.
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done
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Thank you solved as 7C3=35
Date Posted:
3 years ago
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