Questions on Algebra: Combinatorics and Permutations answered by real tutors!

There are two methods for finding this:

Method 1:



and 

Method 2:



Explanation of method 1:

Since "seventeen" is a 9-letter word, there are 9 positions 
to place the letters.

We can choose any 1 of the 9 positions for the "s". 
That's 9C1 ways to place the s. After placing it, that
leaves 8 positions to fill.

We can then choose any 4 of the 8 remaining positions 
for the 4 "e"'s. That's 8C4 ways to place the 4 e's. After 
placing them, that leaves 4 positions to fill.

We can then choose any 1 of the 4 remaining positions 
for the "v". That's 4C1 ways to place the "v". After 
placing it, that leaves 3 positions to fill.

We can then choose any 2 of the 3 remaining positions 
for the 2 "n"'s. That's 3C2 ways to place the 2 "n"'s. After 
placing them, that leaves only 1 position to fill.

So we can chose the position for the "t" in only 1C1 or 1 way.

So the answer using method 1, is



Explanation of Method 2:

First pretend that the 4 e's are all different colors, so that we 
could tell them apart. And also pretend that the 2 n's are different 
colors, so that we can tell them apart too.

That is let's pretend that the word "seventeen" is spelled with
colored letters, like this

seventeen

So there would be 9! or 362880 ways to arrange them if we could 
tell the e's apart and also tell the n's apart.

So one random sample of such an arrangment of "seventeen" is, 
say, this:

nevtesene

Now within this one particular sample arrangement there are 4! ways
we can arrange the colors of the e's and there are 2! ways we can
arrange the colors of the n's. 

So that's 4!2! or 48 ways to move the e's and n's around in that 
one particular sample. And every one of these would be a
case of the arrangement "nevtesene". 

Every sample would be the same, so that means that the 9! or 362880
counts every arrangement 4!2! or 48 times too many, so we must 
divide it by 48 to get the correct answer. So the answer, using 
method 2 is

 

Edwin

Here is a sample choice:

3206

The number that goes where the 3 is can be any of 1, 2, 3, 4, 5, or 6.
0 cannot come first, so we can choose it any of 6 ways.

The number that goes where the 2 is can be 0 or any one of the
5 that was not chosen first.  That's 6 ways.

The number that goes where the 0 can be any of the 5 that weren't
chosen first or second. That's 5 ways.

The number that goes where the 6 is can be any one of the 4 that
weren't chosen 1st, 2nd or 3rd.  That's 4 ways.

The answer is  or  ways. 

Here is a sample choic:

4361

Let's choose the 4th or last digit, where the 1 is, 
first.  It can only be 1, 3, or 5, since the number 
must be odd. That's 3 ways.

Next choose the first digit.  It cannot be 0 or the
digit that was chosen as the 4th digit.  That's
5 ways.

Next choose the second digit.  It can be 0 or any
digit that was not chosen as the 4th digit.  That's
5 ways.

Finally choose the third digit.  It can be any one of
the 4 remaining digits.  

The answer is  or  ways.

There are two ways to solve this.  

First way.

A. The number of ways the 3 can come first.

Sample choice 3256:

There are 6 choices for the digit where the 2 is.
There are 5 choices for the digit where the 5 is.
There are 4 choices for the digit where the 6 is.

That's  or  ways.

B. The number of ways the 3 can come 2nd.

Sample choice 4306:

There are 5 choices for the digit where the 4 is.
There are 5 choices for the digit where the 0 is.
There are 4 choices for the digit where the 6 is.

That's  or  ways.

C. The number of ways the 3 can come 3rd.

This is also  for we could just swap the
2nd and 3rd digits of B.

D. The number of ways the 3 can come 4th.

This is also  for we could just swap the
2nd and 4th digits of B.

That's a total of  or  ways.

Second way.

Find the number which do not contain 3 and then subtract
from the total 720.

Now we are choosing from this set:

{0,1,2,4,5,6}

Here is a sample choice:

2046

The number that goes where the 2 is can be any of 1, 2, 4, 5, or 6.
0 cannot come first, so we can choose it any of 5 ways.

The number that goes where the 0 is can be 0 or any one of the
5 that was not chosen first.  That's 5 ways.

The number that goes where the 4 is can be any of the 4 that weren't
chosen first or second. That's 4 ways.

The number that goes where the 6 is can be any one of the 3 that
weren't chosen 1st, 2nd or 3rd.  That's 3 ways.

The answer is  or  ways. 

So we subtract these which don't contain 3 from the 720 and
get  or .  This is the same answer
we got when doing it the other way.

In order to be divisible by 5, the last digit can only be 0 or 5.

A.  Cases when 0 comes last.

Here is a sample choice:

5240 

There are 6 choices for the digit where the 5 is.
There are 5 choices for the digit where the 2 is.
There are 4 choices for the digit where the 4 is.

That's  or  ways.

B.  Cases when 5 comes last.

Here is a sample choice:

3625

There are 5 choices for the digit where the 3 is,
as it cannot be 0.
There are 5 choices for the digit where the 6 is.
There are 4 choices for the digit where the 2 is.

That's  or  ways.

The total from A and B is

 or  ways.

Edwin

Warning! 
Alan's solution is incorrect because 
he used 10×10 or 100 for the number 
of choices for the side dishes. This 
counts selecting, say, corn first and 
slaw second as different from selecting
slaw first and corn second, when they 
are the same choice.  That's why his
answer is too large.

Edwin's solution:

Number of ways to choose the main course, (which is 5)

AND (TIMES)

Number of ways to choose the side dishes:

  A.  Number of ways to choose two different side dishes, 10C2
  
  OR (PLUS)
  
  B.  Number of ways to choose the same side dish twice, 10

AND (TIMES)

Number of ways to choose the dessert 

That's 









Edwin