Arranging Letters

Date: 11/08/98 at 17:53:06
From: mike
Subject: Discrete math

1) In how many ways can the letters in MISSISSIPPI be arranged?
   Suppose the 2 P's must be separated?

2) In how many ways can the 26 letters in the alphabet be arranged so 
   that the letter b is always to the immediate left of the letter e?
   Suppose we merely require that b be somewhere to the left of e?

3) There are 40 boys and 60 girls.
    In how many ways can a 10-person committee be formed?
    Suppose the committee must contain 5 boys and 6 girls?
    Suppose the committee must contain 6 people of the same gender?

4) A student must answer 8 out of 10 questions in an exam.
    How many choices does the student have?
    How many choices if the first three questions have to be answered?
    How many choices if at least 4 of the first five questions must be   
      answered?

Date: 11/09/98 at 07:25:02
From: Doctor Anthony
Subject: Re: Discrete math

1) There are 11 letters. 

   4 s's   4 i's    2 p's    1  m.

                             11!
Number of arrangements =  ----------  =  34650
                           4! 4! 2!


>Suppose the 2 P's must be separated?

   Consider the 9 other letters arranged in a row.

     - M - i - s - s - i - s - s - i - i -

Including a place at the beginning and end of the row we must select 
two positions to put the p's from the 10 positions available. These 
two positions can be chosen in C(10,2) = 45 ways. The other 9 letters 
could be arranged in 9!/(4! 4!) = 630 ways. So the total number of 
ways is 45 x 630 = 28350 ways.

An alternative method is to stick the two p's together and find 
the number of arrangements when they are together. We have 
10 objects to permute and the number of different arrangements is  
10!/(4! 4!) = 6300

Subtract this from the total of all possible arrangements, 34650, to 
get the number of arrangements with the two p's separated

    34650 - 6300 = 28350

2) Stick the two letters together to form a single object (be). There 
are now 25 objects to permute, and this can be done in 25! ways.

>Suppose we merely require that b be somewhere to the left of e?

We can think globally for this one. There is no reason why b should 
come before e more often than after e when the total of all 26! 
arrangements is considered. So in half the total b will be somewhere 
to the left of e.

The required number is therefore  26!/2

In the following questions I shall be using the formula for the 
number of combinations of r things that can be made from n different 
things.

Use the notation C(n,r) to mean the number of combinations of r things 
that can be made from n different things. The formula for C (n,r) is:

              n!                       10!
  C(n,r) = -------    so   C(10,4) = ------  = 210
           r!(n-r)!                   4! 6!

3)
> There are 40 boys and 60 girls.
> In how many ways can a 10 person committee be formed?

                                               100!
Number of possible committees is C(100,10) =  -------
                                               10! 90!   
         
> Suppose the committee must contain 5 boys and 6 girls?

                                               40!       60!
Number of committees is  C(40,5) x C(60,6) = ------- x ------
                                              5! 35!    6! 54!

> Suppose the committee must contain 6 people of the same gender?

                          40!          60! 
    C(40,6) + C(60,6) = --------  +  -------
                         6! 34!       6! 54!

4)
>A student must answer 8 out of 10 questions in an exam.
> How many choices does the student have?

    C(10,8) = 45

> How many choices if the first three questions have to be answered?

    C(7,5) = 21 

> How many choices if at least 4 of the first five questions must be  
  answered?  

 He could choose 4 out of 5 and then 4 out of another 5

           = C(5,4) x C(5,4) = 5 x 5 =  25

 Or he could choose 5 out of 5 and then 3 out of 5

           = C(5,5) x C(5,3) = 1 x 10 = 10

   The total choice is then 25 + 10 = 35

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/