Math Help please (1 Viewer)

[Cherrybomb56]

Can you guys please explain these two questions to me. Thanks.

 

[Drdusk]

Can you guys please explain these two questions to me. Thanks.

 

[CM_Tutor]

You can also get this result using the identity and equating coefficients of the term in .

 

[Cherrybomb56]

Thanks

 

[Cherrybomb56]

You can also get this result using the identity and equating coefficients of the term in .

Thank you

 

[Velocifire]

How do you write so neat?

 

[CM_Tutor]

Regarding the second question:

(a) 4 digit PINs from 0, 1, 2, ..., 9 with repetition allowed:

Method 1: Since repetition is allowed, the set of all possible PINs is:

{0000, 0001, 0002, 0003, ..., 9999}

The number of all possible PINs is then the cardinality of this set, which is 10,000

Method 2:
The first digit can be chosen as any of the digits from 0 to 9, so there are 10 possible choices.
Similarly for the second digit, and the third, and the fourth.
Thus, # PINs = 10 x 10 x 10 x 10 = 10,000

(b) 4 digit PINs from 0, 1, 2, ..., 9 with repetition prohibited:

Method 1:
The first digit can be chosen as any of the digits from 0 to 9, so there are 10 possible choices.
The second digit can be chosen as any of the digits from 0 to 9 except for the one chosen first, so there are 9 possible choices.
The third digit can be chosen as any of the digits from 0 to 9 except for the ones chosen first or second, so there are 8 possible choices.
The fourth (and last) digit can be chosen as any of the digits from 0 to 9 except for the three already chosen, so there are 7 possible choices.

Thus, # PINs = 10 x 9 x 8 x 7 = 5,040

Method 2:
We know that there are 10,000 possible PINs with no restrictions on repetition, so the number possible without repetition is this number less the number of prohibited cases.

Prohibited Case 1: All four digits the same.
There are ten such PINs (0000, 1111, 2222, ..., 9999) as there are ten possible choices for the digit to repeat.

Prohibited Case 2: Three digits the same.
The digit repeated three times can be chosen in 10 ways.
This repeated digit can be placed into the PIN in 4 ways (XXXY, XXYX, XYXX, or YXXX).
The remaining digit can be chosen in 9 ways.
So, # prohibited three-digit-same PINs = 10 x 4 x 9 = 360.

Prohibited Case 3: Two digits the same.
This case needs care as the remaining two digits could also be the same...

Case 3(a): Two digits the same and the other two different (like 1123)
The digit to be appear twice in the PIN can be chosen in 10 ways.
These identical digits can appear in a PIN in 6 ways (XX.., X.X., X..X, .XX., .X.X, ..XX)
There are 9 ways to choose the next digit, and 8 ways to choose the last digit
So, # prohibited AABC PINs = 10 x 6 x 9 x 8 = 4,320

Case 3(b): Two digits the same and the other two a different pair (like 1122)
The two digits each to appear twice in the PIN can be chosen in ¹⁰C₂ = 45 ways.
There are 6 ways to rearrange an AABB PIN into a different PIN (AABB, ABAB, ABBA, BAAB, BABA, BBAA)
So, # prohibited AABB PINs = 45 x 6 = 270

Total Prohibited PINs = 10 + 360 + 4,320 + 270 = 4,960

So, # allowed 4 digit PINs without replacement = 10,000 - 4,960 = 5,040

Now, see if you can figure out (c) for yourself

 

[Drdusk]

How do you write so neat?

You mean the formatting in the post?

It's called LaTeX a text formatter/editor and it's inbuilt into this website. It's very very helpful for university where your assignments have to look like professional papers.

 

[Velocifire]

No I’m talking about the photo cherrybomb uploaded with her handwriting, so neat!

I know latex hahaahah it’s in my signature

 

[Cherrybomb56]

Regarding the second question:

(a) 4 digit PINs from 0, 1, 2, ..., 9 with repetition allowed:

Method 1: Since repetition is allowed, the set of all possible PINs is:

{0000, 0001, 0002, 0003, ..., 9999}

The number of all possible PINs is then the cardinality of this set, which is 10,000

Method 2:
The first digit can be chosen as any of the digits from 0 to 9, so there are 10 possible choices.
Similarly for the second digit, and the third, and the fourth.
Thus, # PINs = 10 x 10 x 10 x 10 = 10,000

(b) 4 digit PINs from 0, 1, 2, ..., 9 with repetition prohibited:

Method 1:
The first digit can be chosen as any of the digits from 0 to 9, so there are 10 possible choices.
The second digit can be chosen as any of the digits from 0 to 9 except for the one chosen first, so there are 9 possible choices.
The third digit can be chosen as any of the digits from 0 to 9 except for the ones chosen first or second, so there are 8 possible choices.
The fourth (and last) digit can be chosen as any of the digits from 0 to 9 except for the three already chosen, so there are 7 possible choices.

Thus, # PINs = 10 x 9 x 8 x 7 = 5,040

Method 2:
We know that there are 10,000 possible PINs with no restrictions on repetition, so the number possible without repetition is this number less the number of prohibited cases.

Prohibited Case 1: All four digits the same.
There are ten such PINs (0000, 1111, 2222, ..., 9999) as there are ten possible choices for the digit to repeat.

Prohibited Case 2: Three digits the same.
The digit repeated three times can be chosen in 10 ways.
This repeated digit can be placed into the PIN in 4 ways (XXXY, XXYX, XYXX, or YXXX).
The remaining digit can be chosen in 9 ways.
So, # prohibited three-digit-same PINs = 10 x 4 x 9 = 360.

Prohibited Case 3: Two digits the same.
This case needs care as the remaining two digits could also be the same...

Case 3(a): Two digits the same and the other two different (like 1123)
The digit to be appear twice in the PIN can be chosen in 10 ways.
These identical digits can appear in a PIN in 6 ways (XX.., X.X., X..X, .XX., .X.X, ..XX)
There are 9 ways to choose the next digit, and 8 ways to choose the last digit
So, # prohibited AABC PINs = 10 x 6 x 9 x 8 = 4,320

Case 3(b): Two digits the same and the other two a different pair (like 1122)
The two digits each to appear twice in the PIN can be chosen in ¹⁰C₂ = 45 ways.
There are 6 ways to rearrange an AABB PIN into a different PIN (AABB, ABAB, ABBA, BAAB, BABA, BBAA)
So, # prohibited AABB PINs = 45 x 6 = 270

Total Prohibited PINs = 10 + 360 + 4,320 + 270 = 4,960

So, # allowed 4 digit PINs without replacement = 10,000 - 4,960 = 5,040

Now, see if you can figure out (c) for yourself

Thanks

 

[Cherrybomb56]

No I’m talking about the photo cherrybomb uploaded with her handwriting, so neat!

I know latex hahaahah it’s in my signature

it's really not that neat but thanks.