Permutation and Combination

• Audi

  1 May 11, 18:36

  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

  1 May 11, 18:49

  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

  1 May 11, 18:54

  Originally posted by wee_ws:

  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

  1 May 11, 19:14

  Hi,

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

  Cheers,
  Wen Shih

• frekiwang

  2 May 11, 18:17

  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

  2 May 11, 18:58

  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

  24 Jul 11, 01:27

  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 (45x44x43x42x41x40x39) 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!! :)
• frekiwang

  24 Jul 11, 20:46

  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

  26 Jul 11, 06:50

  Hi,

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

  Cheers,
  Wen Shih

• frekiwang

  26 Jul 11, 08:32

  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

  26 Jul 11, 14:06

  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

  27 Jul 11, 11:08

  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

  30 Jul 11, 17:34

  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

  30 Jul 11, 21:55

  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.

   

  Q2

  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.