how many 6-digit numbers we can get from the digits of 14145566?

Question

It can, and will be repeated.

I was thinking of $$\bar{A}_n^k = n^k,$$

but it seems to be too easy.

Should it not be multiplication like:

$$\bar{A}_6^2 \cdot \bar{A}_6^2 \cdot \bar{A}_6^2 $$

because we only need 6-digits numbers and every digit we see only two times in the original number?

Answer 1

many ways to do this but the possibility of double digits really throws in a monkey wrench. But we can be careful to not fall for traps.

We need to pick six digits from 8 digits. So there are two digits we will not use. They can either be the same or different. So

So Case 1: The ones we don't pick are the same... or in other words... the ones we do pick are three pairs so .... we make six digit numbers for 3 pairs of digits.

There are ${4 \choose 3}=4$ ways to choose the $3$ pairs of digits. There are ${6\choose 2} = 15$ ways to place the first pair and ${4\choose 2} = 6$ ways to place the second pair and ${2\choose 2} = 1$ ways to place the third pair.

So there are ${4 \choose 3}{6\choose 2}{4\choose 2}{2\choose 2}= 4*15*6 = 360$ ways to make a six digit number with 3 pairs of digits.

Case 2: The ones we don't pick are different... or in other words... the ones we do pick are two pairs of digits and two singletons so ....We have a six digit number with 2 pairs of digits and 2 singletons.

There are ${4\choose 2} = 6$ ways to choose which of the digits we use as pairs and which we use as singletons. There are ${6\choose 2} = 15$ ways to place the first pair and ${4\choose 2} = 6$ ways to place the second pair and $2!=1$ ways to place the two singletons.

So there are ${4\choose 2}{6\choose 2}{4\choose 2}2! = 6*15*6*2= 1080$ ways to make a six digit number with 2 pairs and 2 singletons.

So there are $360 + 1080 = 1440$ total six digit numbers.

There are probably other strategies we could use, but this is the one that comes clearest to my brain.

Answer 2

Hint: How many ways can we pick the numbers $1,1,4,4,5,5,6 \text{ and }6$ out of a bag?

Answer 3

The six digit no. may have either

1) only 3 types of digits:

(Each digit will repeat 2 times)

Those 3 digits can be chosen in $4$ ways. And after chosing the digits the ways of arrangement will be:

$6!/(2!.2!.2!)$

Hence total ways $=4(6!/(2!.2!.2!))$

           $OR$

2) all 4 types of digits:

(only 2 digits will appear twice and 2 will appear once)

Ways to chose the 2 repeating digits $=6$

Ways to arrange(after chosing digits) $=6!/(2!.2!)

Total ways $=6(6!/(2!.2!))$

Taking both the cases together

( adding both the totals)

Total ways $= 2.6!= 1440$