Permutations (2 Viewers)

[VBN2470]

Nup, there are no factorials or perms/combs here. If the coins and the boxes are both distinguishable, then each coin has 4 possibilities, each leading to a different result. So the number of cases if we allow a box to be empty is 4¹⁰.

If we want at least one coin in each box, we have to subtract cases from this. I am getting 4¹⁰ - 4×3¹⁰ + 6×2¹⁰ - 8. But I'm not too sure about this and am prepared for a correction.

Wouldn't it still be 10! ways to arrange the coins if the boxes were still distinct?

 

[braintic]

Wouldn't it still be 10! ways to arrange the coins if the boxes were still distinct?

Factorials come into play only when repetition is not allowed, hence the 'reduce by one before multiplying' effect. Putting the first coin into box A doesn't prevent you from also putting the second coin into that box, so repetition is definitely allowed.

In any case, the first choice is 'which box will I put this coin into', so there are never 10 options, only 4 at a time. We are assigning coins to boxes, not boxes to coins. The only way it would be 10! is if you have 10 distinct coins AND 10 distinct boxes, and want to place EXACTLY one coin in each box, not AT LEAST one.

 

[VBN2470]

OK, then how would you do the following question (from Terry Lee 3U Ch7 Q24(b))?

In how many ways can 9 different rings be arranged on the 4 fingers (excluding the thumb), assuming that no fingers can be empty?

Answer says 8C3 X 9! = 20 321 280, where they have arranged the 9 different rings and inserted 3 separators in 8 possible 'gaps'.

This is just like the coin example, except now we adjusted it so that the coins are distinct and not identical to each other, so I want to know if the same method applies.

 

[braintic]

OK, then how would you do the following question (from Terry Lee 3U Ch7 Q24(b))?

In how many ways can 9 different rings be arranged on the 4 fingers (excluding the thumb), assuming that no fingers can be empty?

Answer says 8C3 X 9! = 20 321 280, where they have arranged the 9 different rings and inserted 3 separators in 8 possible 'gaps'.

This is just like the coin example, except now we adjusted it so that the coins are distinct and not identical to each other, so I want to know if the same method applies.

That depends. Are they allowing for different orderings of the rings on each finger? If so, it is not the same question. There is no ordering of the coins within each box.

If they are not doing that, then their answer MUST be wrong. The MAXIMUM number of ways you can allocate the rings without ordering is 4^9, assuming there are absolutely no restrictions, and this is only 262 144.

 

[VBN2470]

Ah, so what I did was assume that there was some sort of ordering of the coins within each of the boxes as well, which led me to believe they could be arranged in 10! ways, whereas we are only considering the ways in the coins can be assigned to the box (repitition allowed), hence 4^10 ways?

Far out, although Permutations and Combinations topics is intriguing at times, it can also be a massive mindf**k as well, especially doing it two years after the HSC.

 

[braintic]

it can also be a massive mindf**k as well

Any f**k is a good f**k !