Find the number of ways the cake can be shared among two people

[chwala]

Homework Statement

This is from an A level past paper question;

i. A plate of cake holds $12$ different cakes. Find the number of ways these cakes can be shared between Alex and James if each receives an odd number of cakes.

Relevant Equations

Stats

I went through the question; find the mark scheme here;

Well i can follow the mark scheme steps but i need some clarity or rather insight. The $12$ cakes are being shared to the two persons. In my understanding the odd numbers are;

$[1,3,5,7,9,11]$ Now this means that they may each get $1$ cake in $1 ×11C_1$ ways or alternatively $^{12}C_1×2$persons...

...are they not supposed to have $2048 ×2$?

Supposing it was $13$ people instead of $2$ ...Would the steps still be the same as shown on the markscheme? A bit confusing...

your insight appreciated...

 

[pasmith]

Homework Statement:: This is a past paper question;

A plate of cake holds $12$ different cakes. Find the number of ways these cakes can be shared between Alex and James if each receives an odd number of cakes.
Relevant Equations:: Stats

I went through the question; find the mark scheme here;

View attachment 319104

Well i can follow the mark scheme steps but i need some clarity or rather insight. The $12$ cakes are being shared to the two persons. In my understanding the odd numbers are;

$[1,3,5,7,9,11]$ Now this means that they may each get $1$ cake in $1 ×11C_1$ ways or alternatively $12C_1×2$persons...

My understanding (which is consistent with the mark scheme) is that they can't each get one cake: If Alex gets one cake then James gets the remaining 11. There are ${}^{12}C_{1} = {}^{12}C_{11} = 12$ ways to achieve this.

 

[chwala]

My understanding (which is consistent with the mark scheme) is that they can't each get one cake: If Alex gets one cake then James gets the remaining 11. There are ${}^{12}C_{1} = {}^{12}C_{11} = 12$ ways to achieve this.

Ok, that's clear now, maybe i was not interpreting the language correctly...

... I get it now ${}^{12}C_{3} = {}^{12}C_{9} = 220$ ways ...

 

[chwala]

This is the second part of the question; i do not seem to get it.

ii. Another plate holds $7$ cup cakes, each with a different colour icing, and $4$ Brownies, each of a different size. Find the number of different ways these $11$ cakes can be arranged in a row if no Brownie is next to another Brownie.

Find the solution here;

In my understanding, they clamped the Brownies together i.e

1

2

3

4

5

6

7

Brownie

Its clear to me that the $7$ cup cakes can be arranged in $7!$ ways no problem there... now when it come to the Brownies, i do not seem to understand the $^8P_4$ ways. Does the $4$ apply to Brownies only or all....

Unless the reasoning is like this;

B

B

B

B

but each cell having the Brownie can arranged in ##4!# ways...secondly supposing we amend the question to all Brownies have to be next to each other, then how would this look like?

cheers

 

[chwala]

My other reasoning on this; since we do not want the brownies to be next to each other then the only possibility would be to have barriers in between i.e the cup cakes, that is;

B

C

B

C

B

C

B

C

The cup cakes can be arranged like earlier stated in $7!$ ways...In whichever way we arrange them it does not matter as long as their sum total (cup cakes) =7.

For e.g;

B

4Cup Cakes

B

C

B

C

B

C

or

B

2 Cup cakes

B

1 Cupcake

B

3 Cup cakes

B

1 Cupcake

Now we have $8$ elements implying that the brownies can be arranged in $^8P_4$ ways...giving us the desired; $7! ×^8P_4=8467200$ ways.

Cheers! Bingo!