This Permutation question kept me up all night. Kindly assist

T

Tenotea

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The question goes thus:

Question: 5-digits numbers are formed from digits 4, 5, 6, 7 and 8.
(a) How many of such numbers can be formed if repetition of digits is
(i) allowed? (ii) not allowed?
(b) How many of the numbers are odd, if repetition of digits is not allowed?

I was able to solve the first part of the question and the solution is below. But the second part is what actually troubled me.

SOLUTION
(a)
(i) If repetition of digits is allowed, nr number of digits can be formed.
and n=5, r=5
Therefore, 55=3125 number of digits can be formed if repetition of digits is allowed.

(ii) If repetition of digits is not allowed, n! number of digits can be formed and n=5
Therefore, 5!= 5*4*3*2*1=120 number of digits can be formed if repetition of digits is not allowed.
 
Use the multiplication principle:

___ x ___ x ___ x ___ x ___

Which position (1st to 5th?) has the restriction placed on it?
How many ways can this position be filled?
After that place is filled how many ways can the other places be filled?

Show what you can do and we'll see how well you understand.
 
Alternatively, seeing that you have found that there are 120 numbers that can be formed, what fraction of those will end in a 4, 5, 6, 7 or 8?
So what fraction will be odd?
So how many are odd?
 
The solution is thus:
(b) 2 of the given 5 digits are odd.
Therefore, 2/5 of 120 = 48 numbers are odd.
Thank you.
:);)
 
Would you like me to explain the first method I suggested It may help with other problems.
 
That method is used in getting the number of digits that can be formed if repetition is not allowed right?
 
"(i) If repetition of digits is allowed, nr number of digits can be formed.
and n=5, r=5
Therefore, 55=3125 number of digits can be formed if repetition of digits is allowed.

(ii) If repetition of digits is not allowed, n! number of digits can be formed and n=5
Therefore, 5!= 5*4*3*2*1=120 number of digits can be formed if repetition of digits is not allowed."

Your solutions above seem to rely on a "formula". That's fine if the questions are standard but causes issues if there's a bit of a twist in the question as in (b).

The "formulas" you've learnt rely on the "multiplication principle".

In all cases there are 5 spaces to fill

Case (I): ___ x ___ x ___ x ___ x ___
You have 5 position to fill.
Here repetition is allowed so you have 5 choices to fill the first position. Put a 5 in there.
Then you have 5 choices to fill position 2. Put a 5 in there. Same for other positions.
This gives 5x5x5x5x5 = 55.

Case (2): ___ x ___ x ___ x ___ x ___
Again you have 5 positions to fill.
You have 5 choices to fill the first position.
Once you've filled it there are only 4 choices for the second position (since no repetition is allowed)
Then there are 3 choices for the 3rd; 2 choices for the 4th and 1 choice for the 5th.
This gives 5x4x3x2x1 = 5!

Case (3): ___ x ___ x ___ x ___ x ___ ODD and no repetitions.
"Odd" means the last digit must be 5 or 7, that's 2 choices. Put a 2 in the last position (always deal with the restriction first).
Once you've filled that position there are 4 ways of filling the first; 3 for the 2nd; 2 for the 3rd; 1 for the 4th. (The last is already filled.)
This gives 4x3x2x1x2 =48

Aim to understand the process; the "formulas" wont always help.
 
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