# Maths - Baker's Dozen

Extracts from this document...

Introduction

Baker’s Dozen Coursework

A baker had laid out three Chelsea Buns (C) and three Bath Buns. His wife decided that they would look better if all of the Chelsea Buns were moved to the left hand side, and all of the Bath Buns to the right hand side. In order to achieve this she would have to swap a bun with its neighbour in the smallest possible number of moves. This means swapping:

To

Starting by swapping the Chelsea Buns first, bringing them to the left hand side would be illogical as it would mean swapping a Chelsea Bun with a Chelsea Bun. So, in this case we will start with moving the Bath Buns first.

We start with one Bath Bun on the RHS, and a Chelsea bun on the LHS.

Start

We are now bringing the Bath Buns over to the RHS, forcing the Chelsea Buns to the LHS.

Swap 1

We have collected two Baths Buns to the RHS, leaving one isolated.

Swap 2

The final swap has been made leaving all the Baths Buns on the RHS and the Chelsea Buns on the LHS.

Swap 3

After investigating with other possible solutions I have deducted that the smallest number of swaps needed to sort out three Chelsea Buns and three Bath Buns is three, as shown above.

Middle

½+ (-½) +c=0

c=0

We can then deduct that:

Tn=½n2+ (-½) n+0

Tn=½n2-½n

We can simplify this to become

Sn=n (n-1), where S is the number of swaps.

2

If this formula is correct for n then it should be correct for all other numbers/ letters. To prove this I shall replace the n with a K which is a constant. The formula would then become

Sk = k (k–1)

2

Now I must prove that the formula can work where n=k+1

RHS: = (k+1) (k+1-1)

2

= k (k+1)

2

We have proven the right hand side of the formula. The left hand side must now match.

LHS: = Sk+1

= Sk + (k)

= k (k-1) + k

2 1

= k (k-1) + 2k

2

= k2 –k+2k

2

= k2+ k

2

= k (k+1)

2

The left hand side is now also proven. Both sides are the same as they equal k (k+1). Therefore the formula is correct.

2

I will now check the formula by inserting a number which we know the answer to (4 of each bun, results in 6 swaps).

Sn=n (n-1) Substitute the numbers in S4= 4x (4-1)

2 2

= 4x 3

2

=12

2

= 6 swaps.

We can see from our table of results that there are six swaps when there are four of each bun meaning that this formula is definitely correct.

In the third question I will add more variations.

Conclusion

Tn = 1.5n2 – 1.5n

Tn = 1.5 x 32 – 1.5 x 3

Tn = 13.5 - 4.5

Tn = 9

Nine swaps are correct for three of each bun, which has been shown above by diagrams and in the table. Therefore this formula is correct and has been proven.

The second variation I will use is that the Chelsea Buns and Bath Buns are arranged in a different order that is:

It will have to be rearranged to form:

One pair of each bun:

No swaps have to be made as this is already in order.

Two pairs of each bun:

Start

Swap 1

Swap 2

Swap 3

Swap 4

Only four moves are needed to sort out eight buns.

3 pairs of each bun:

Start

Swap 1

Swap 2

Swap 3

Swap 4

Swap 5

Swap 6

Swap 7

Swap 8

Swap 9

Swap 10

Swap 11

Swap 12

There are twelve swaps needed when there are three pairs of each bun.

We have enough information to draw up another table.

Number of Each Pair | Number of All Buns | Number of Swaps |

1 | 4 | 0 |

2 | 8 | 4 |

3 | 12 | 12 |

4 | 16 | 24 |

We must use a linear sequence. The common difference is 4 as we can see below.

0 4 12 24

4 8 12

4 4

To work out the formula we must use the rule Tn = an2 + bn + c.

2a = 4

a = 2

3a+ b= 4

3 x 2 +b = 4

6 + b = 4

b= -2

a + b + c = 0

2 + (-2) + c = 0

1 + c = 0

c = 0

Tn = 2n2 – 2n

Substitution can be used once again.

Tn = 2n2 – 2n Tn = 2n² - 2n

Tn = 2 x 4 – (2 x 2) Tn = 2 x 16 – (2 x 4)

Tn = 8 – 4 Tn = 32 - 8

Tn = 4 Tn = 24 Four is also right.

This is correct as for 2 pairs of buns there are four swaps.

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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