Can someone please explain this to me? Thanks in advance!

[Muppers3262]

I know that the digits 1, 2, 3, and 4 can be arranged 24 different ways when forming four-digit numbers, but how do I found out what position 3142 is in when the 24 four-digit numbers are listed in order from smallest to largest? :?

I did do this out by lisiting the numbers and think that 3142, is the 14th number when placing in order from smallest to largest.

Please explain if there is an actual process, besides listing out the numbers in which I can find the answer. Thank you!

 

[zerohour]

I don't think there is a formula in this case. However by logical deduction you should get the answer without listing.
We have to list from smallest to largest
So with 1 as the first number we can have 6 combinations with 2,3 and 4 in the second, third and fourth place. (3 X 2 = 6)
Again with 2 as the first number we get 6 more combinations. That makes it a total of 12.
Then with three as the first number and one as the second, the thirteenth number would be 3124.
Fouteenth would be 3142!

 

[TchrWill]

I know that the digits 1, 2, 3, and 4 can be arranged 24 different ways when forming four-digit numbers, but how do I found out what position 3142 is in when the 24 four-digit numbers are listed in order from smallest to largest?

I did do this out by lisiting the numbers and think that 3142, is the 14th number when placing in order from smallest to largest.

Please explain if there is an actual process, besides listing out the numbers in which I can find the answer.

Not sure this meets the requirements of a formula, but it gets your answer.

Looking at the list of the first 6, we have 1234, 1243, 1324, 1342, 1423, and 1432

Each successive group starts with 2, 3 and 4.

The number youare seeking the position of is 3142 indicating that the position of the number in the total array of 24 numbers is in the 3rd group, its position being indicated by 12 + N.

The second number in those of group 3 are 1, 2 and 4.

The 3rd and 4th numbers must be 2 of the remaining 2 numbers in order.

N must therefore be either 1, 2 or 4.2nd number being 4.

You know from the first set of 6 numbers that there are 2 numbers with the 2nd number being 1, 2 numbers with the 2nd number being 2 and 2 numbers with the 2nd number being 4.

Since your 2nd number is 1, you immediately know that the position of your number in the list is 12 + 1 or 12 + 2.

The last 2 numbers must be in ascending order telling you that tthe position of your number is 12 + 2 = 14, since the 4 comes before the 2.

This can no doubt be put into some sort of equation wich I will let you tacke.

 

[soroban]

Re: Can someone please explain this to me? Thanks in advance

Hello, Muppers3262!

I have an answer for the "reverse" problem:
. . Given the position of the number, find the number.

I know that the digits 1, 2, 3, and 4 can be arranged 24 different ways when forming four-digit numbers,
but how do I found out what position 3142 is in
when the 24 four-digit numbers are listed in order from smallest to largest?

In 1961, a high school student, Dale Kozniuk, found this procedure.

<u>Example</u>

Given the digits {1,2,3,4,5}, .5! = 120 different five-digit numbers can be formed.
Arranging them in increasing order, what is the 37^th number on ths list?

Write 37 as a "factorial polynomial".
. . 37 .= .1(4!) + 2(3!) + 0(2!) + 0(1!) + 1(0!)

That is, divide 37 by successively smaller factorials.

. . Important: .37 .= .1(4!) + 2(3!) + 1
. . . . The final 1 could be written: .0(2!) + 1(1!)
. . . . However, the rule is: always include the 0!.

. . Although 2 = 1(2!), it must be written: .2 .= .0(2!) + 1(1!) + 1(0!)
. . . . That is, we must "spread it out as thin as possible."


Note the coefficients of the "polynomial: .1, 2, 0, 0, 1

Add 1 to all coefficients but the last: .2, 3, 1, 1, 1

These are ordinal numbers: .2^nd, 3^rd, 1^st, 1^st, 1^st.

We begin with the first permuation "12345" and select the digit in the indicated position.

position . . permutation . . digit

. 2nd . . . . . 1 2 3 4 5 . . . . . 2

. 3rd . . . . . 1 . .3 4 5 . . . . . .4

. 1st . . . . . 1 . .3 . .5 . . . . . .1

. 1st . . . . . . . . 3 . .5 . . . . . .3

. 1st . . . . . . . . . . . 5 . . . . . .5

Therefore, the 37^th number on the list is 24135.


This result can be checked by a "reverse problem".

Writing the 120 five-digit in <u>decreasing</u> order,
. . What is the: .120 - 37 + 1 .= .84^th number?

84 .= .3(4!) + 1(3!) + 2(2!) + 1(2!) + 1(0!)

The coefficients are: .3, 1, 2, 1, 1

The ordinals are: .4^th, 2^nd, 3^rd, 2<syp>nd</sup>, 1^st

position . . permutation . . digit

. 4th . . . . . 5 4 3 2 1 . . . . . 2

. 2nd . . . . .5 4 3 . .1 . . . . . 4

. 3rd . . . . . 5 . 3 . . 1 . . . . . 1

. 2nd . . . . .5 . 3 . . . . . . . . .3

. 1st . . . . . 5 . . . . . . . . . . . 5

Therefore, the 84^th number "from the other end" is: .24135.