Permutation and Combination

Audi SG

Find the number of ways of forming a word using all the letters from the word MATHEMATICAL in such a way that

the 3 "A"s are never next to each other.

Firstly I found the total number of ways of arranging WITHOUT restrictions which is 12!/ (2! x 3! x 2!) = 19 958 400

Then I find the total no. of ways when the 3 As are ALWAYS next to each other and I got 907 200.

Subtracting, I go the answer as 19 051 200. But the suggested answer is 10 886 400. It seems that they subtracted by 9 072 000 and not 907 200. Did i go wrong anywhere?

wee_ws SG

Hi,

It is easier to use the direct approach, consisting of two steps:

1. Arrange the nine other letters.

2. Slot the three As in between.

Thanks.

Cheers,
Wen Shih

Audi SG

Hi,

It is easier to use the direct approach, consisting of two steps:

1. Arrange the nine other letters.

2. Slot the three As in between.

Thanks.

Cheers,
Wen Shih

Yep, I tried that as alternative method but I got the same answer. So I was wondering if I had erred somewhere.

Thanks

wee_ws SG

Hi,

I got the answer, so perhaps you should relook at your working. Thanks.

Cheers,
Wen Shih

frekiwang SG

answer is not equal to “without restriction”-“all As are together”, since you also have to substract the case for which “2As are together and the other is not”

wee_ws SG

Hi,

If a direct method is possible, apply it.

The mutual exclusion principle requires one to have a very good idea of the different cases. Only use it if we are very sure that there is only one situation for the complementary event.

Thanks.

Cheers,
Wen Shih

chisinau SG

Hello,

I have 2 questions here:

Q1: What is the total number of ways to allocate 12 classes into 6 houses: Green, Yellow, Red, Blue, Orange, Purple if each house has the same number of classes?

Q2: What is the total number of combinations in a TOTO draw (ie. 7 numbers out of 45, additional number doesn’t apply)

It should be 45C7 right, since the order of the numbers doesn’t matter. But on the hand I was thinking 45P7 (45×44x43×42x41×40x39) makes sense too – because for the first number, I have 45 choices, for the 2nd number, I have 44 choices, so on and so forth. I can’t the flaw in my logic..

Any help is much appreciated!! Smile

frekiwang SG

Q1. 12C2 x 10C2 x 8C2 x 6C2 x 4C2 x 2C2=.....

Q2. 45C7 is correct, 45P7 is not. The difference is whether 'Order is important'

If 1 2 3 4 5 6 7 and 7 6 5 4 3 2 1 are considered as different cases, then P should be used, otherwise C should be used. As for toto, C is clearly the choice.

You are welcome to clarify if you still have doubts

wee_ws SG

Hi,

There is a need to deal with each house having the same number of classes. Thanks.

Cheers,
Wen Shih

frekiwang SG

There should be no problem with my answer.

12C2 means 'no. of ways of choosing 2 classes from 12 and put them into the Green house', 10C2 means 'no of ways of choosing 2 classes from the remaining 10 and put them into the Yellow House' and etc.

wee_ws SG

Hi,

Yes, indeed, since the problem is assigning distinct objects into distinct groups.

If we are assigning 12 classes into 6 groups of size 2, then we will then need to deal with each group having the same number of classes.

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

I am happy to share some thoughts about counting:

wenshih.files.wordpress.com/2010/02/counting-the-number-of-possible-allocations.pdf

Thanks!

Cheers,
Wen Shih

J2011 SG

Hi,

Q1: A 4-storey house is to be painted by 6 different colours such that each storey is painted in one colour. How many ways are there to paint the house?

Q2: There are five flavors of ice-cream: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. How many variations will be there?

Thanks.

frekiwang SG

Q1.

6P4 = 360 ways

The problem is to 'choose' 4 colours out of 6 and the order of painting 'is important', therefore P is used.

5C3 = 10ways.

The problem is to 'choose' 3 flavours out of 5 and the order of mixing them 'is not important', therefore C is used.