I get it now.

Thanks ...I can explain to my gal now.

## Q&A - P3 Math

### Re: Q&A - P3 Math

The seats, from left to right lets, we label them as S1, S2, S3, S4 and S5love1001 wrote:Hi, can someone help me with this combination type question?

Alice, Ben, Cathy, Daniel and Elle are seated in a row.

Alice can sit either on the extreme left or right.

Cathy must sit in the center.

How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?

Alice can sit either S1 or S5. That poses two scenario to calculate the combinations.

Senario 1, Alice took S1,

S1 - 1 option, i.e Alice

S2 - 3 option, from Ben, Daniel and Elle

S3 - 1 option, Cathy has to sit here

S4 - 2 option. after S2 is taken, we are left to 2 choices.

S5 - 1 option. After S2 and S4 is taken, we are left with 1 choice.

Total combination for Scenario 1,

1 X 3 X 1 X 2 X 1 = 6

Scenario 2,

We fix Alice at S5

S1 - 3 options, from either Ben, Daniel and Elle

S2 - 2 options. After S1 is chosen, we are left to 2 choices

S3 - 1 option. Only Cathy sits here

S4 - 1 option, After S1 and S2 is taken, we are left with one choice

S5 - 1 option, That is for Alice only

Total combination for scenario 2

3 X 2 X 1 x 1 x1 = 6 combination

Total combination = Combination for Scenario 1 + Combination for Scenario 2

6 + 6 = 12 ######

### Re: Q&A - P3 Math

Let's do it another way. Instead finding the combination via the chairs, we find the how many choices does one individual has.Ender wrote:The seats, from left to right lets, we label them as S1, S2, S3, S4 and S5love1001 wrote:Hi, can someone help me with this combination type question?

Alice, Ben, Cathy, Daniel and Elle are seated in a row.

Alice can sit either on the extreme left or right.

Cathy must sit in the center.

How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?

Alice can sit either S1 or S5. That poses two scenario to calculate the combinations.

Senario 1, Alice took S1,

S1 - 1 option, i.e Alice

S2 - 3 option, from Ben, Daniel and Elle

S3 - 1 option, Cathy has to sit here

S4 - 2 option. after S2 is taken, we are left to 2 choices.

S5 - 1 option. After S2 and S4 is taken, we are left with 1 choice.

Total combination for Scenario 1,

1 X 3 X 1 X 2 X 1 = 6

Scenario 2,

We fix Alice at S5

S1 - 3 options, from either Ben, Daniel and Elle

S2 - 2 options. After S1 is chosen, we are left to 2 choices

S3 - 1 option. Only Cathy sits here

S4 - 1 option, After S1 and S2 is taken, we are left with one choice

S5 - 1 option, That is for Alice only

Total combination for scenario 2

3 X 2 X 1 x 1 x1 = 6 combination

Total combination = Combination for Scenario 1 + Combination for Scenario 2

6 + 6 = 12 ######

We start with the one with the most restriction, i.e Cathy and then work ourselves to those with more choices.

Cathy - 1 choice. She has to take S3

Alice - 2 choices . She can choose either S1 or S5

Ben - 3 choices, after Alice selects either S1 or S5, Ben is left with only 3 choices

Daniel - 2 choices. After all the above have chosen, Daniel has two chairs to choose.

Elle - 1 choice . By now 4 chairs have been chosen. Left one choice.

Total combination = 1 X 2 X 3 X 2 X 1 = 12 combinations ###

Last edited by Ender on Sat Oct 08, 2016 3:50 pm, edited 1 time in total.

### Re: Q&A - P3 Math

Can go by method of listing:love1001 wrote:Hi, can someone help me with this combination type question?

Alice, Ben, Cathy, Daniel and Elle are seated in a row.

Alice can sit either on the extreme left or right.

Cathy must sit in the center.

How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?

We commonly see similar question with 3 people (X Y and Z) and most P3 are taught to solve by listing.

XYZ XZY YXZ YZX ZXY ZYX (6 ways, "basic" list)

So for your question: we know C must be centre. A is either on extreme left or right.

Suppose A is on extreme left like this:

**[A ? C ? ?]**

Then remaining B, D and E have be like the above "XYZ" case, 6 ways to fill the remaining 3 seats.

*Notes (just for understanding):

BDE BED DBE DEB EBD EDB (just need to understand this basic list of 6 choices)

This arrangement is the same as below, since "_" are fixed positions taken up by A and C.

_B_DE / _B_ED / _D_BE / _D_EB / _E_BD / _E_DB (same basic list)

But A can also switch to the extreme right like this:

**[? ? C ? A]**

Then in the same way, there are another 6 ways to for B, D and E to fill the remaining 3 seats.

*Notes (not necessary to list again)

BD_E_ / BE_D_ / DB_E_ / DE_B_ / EB_D_ / ED_B_ (same basic list)

Total is 6+6=12 ways#

- oliviachng
- YellowBelt
**Posts:**21**Joined:**Thu Nov 10,

### Re: Q&A - P3 Math

Bunny27 wrote:Hi I need help on this question,

Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?

Thank you

Just have to draw a simple diagram

4 units => 300 + 252 = 552

1 unit => 552/4 = 138

sufia only has one unit in the end, so the answer is 138