Lesson A confusing combinatorial problem on repeating digits in numbers

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<H2>A confusing combinatorial problem on repeating digits in numbers</H2>

<H3>Problem 1</H3>What is the number of different five-digit numbers that can be formed 
from the set &nbsp;&nbsp;S = {2,3,4,5,6} &nbsp;&nbsp;such that one digit is repeated twice and another digit is repeated twice.

<B>Solution</B>

In &nbsp;Mathematics, &nbsp;the words &nbsp;" digit is repeated twice " &nbsp;mean the same as &nbsp;" digit is used twice ".

<pre>
           <U>Solve step by step</U>

(a)  Let's consider more simple problem first:

Given three different symbols A, B and C as an alphabet, 
         how many different 5-symbol words can be formed
         such that symbol A is used twice, symbol B is used twice 
         and symbol C is used once?

For this problem, we can select a position for A in  {{{C[5]^2}}} = 10 ways;
                                         a position for B in  {{{C[3]^2}}} = 3 ways;
                                         then C occupies the remaining 5th position, at no choice.

So, for this problems, there are 10*3 = 30 different words.

This outcome is the same as if you use well known formula for repeating letters

{{{5!/((2!)*2!))}}} = {{{120/(2*2)}}} = 120/4}}} = 30.

(b)  Now, returning to the given problem, we can assign any of 5 possible digit to A;
                                                        any of remaining 4 digits to B,
                                                    and any of remaining 3 digits to C.

It gives 5*4*3 = 60 different choices for assigning; but for us, the pairs 
     (A=2,B=3) and (B=2,A=3) lead to identical undistinguishable numbers.

Therefore, this number 60 we should divide by 2 and take 60/2 = 30.

getting the  <U>ANSWER</U>  30 * 30 = 900.
</pre>

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