I'm not a mathematician. I appreciate mathematicians. I love mathematicians.

But anyhow, 6! seems right to me. And 6!=720

Yes. (I think so too... 1/720 for all of then right).

There's no way to get just five of them right. Thus there are 0/720 ways to get exactly five right.

No.

I don't think that logic works. I try to just use logic I know works.

To get four right, you must have two wrong. It's like having a row of six marbles, four of them white and two of them black. A white marble represents correct placement while a black marble represents incorrect placement.

To have two black,
if the first marble is black, then there are five ways to have another marble be black. And that's it for the first marble.
if the second marble is black, then there are four ways to have another marble (aside from the first) be black. And that's it for the second marble.
if the third marble is black, then there are three ways to have another marble (aside from the first and second) be black. And that's it for the third marble.
if the fourth marble is black, then there are two ways to have another marble (aside from the first, second and third) be black. And that's it for the fourth marble.
if the fifth marble is black, then there is one ways to have another marble (aside from the first, second, third, and fourth) be black. And that's it for the fifth marble.
We've already counted all the ways the sixth marble can be black.

5+4+3+2+1=15. That's the number of ways 2 marbles can be black. Thus that's the number of ways two answers can be exchanged and thus the number of ways exactly four answers can be correct

I don't think that logic works.

For exactly three answers to be wrong, first we choose the three answers that will be wrong, then choose how many ways they can be wrong. Let's do the second part first. If the order is ABCDEF, and we choose ABC to be wrong, there are six (3!) ways to arrange ABC (ABC, ACB, BCA, BAC, CAB, CBA) but it's immediately obvious there are only two arrangements with all three out of place).
Now we just have to decide how many ways we can arrange three white and three black marbles in a row.
000111
001011
001101
001110
010011
010101
010110
011001
011010
011100
100011
100101
100110
101001
101010
101100
110001
110010
110100
111000
If I counted correctly, there are 20 of these. And each one has exactly three misplaced twice.
So 20*2=40. There are 40 ways to have exactly three misplaced (and thus exactly three correctly placed) answers.

To have two correctly placed, consider again black and white marbles. There are 15 ways to arrange 2 white and 4 black marbles. But now when we look at the four black marbles that are misplaced, think of ABCD. 4!=24 possibilities. Here they all are:

ABCD no
ABDC no
ACBD no
ACDB no
ADBC no
ADCB no
BACD no
BADC yes
BCAD no
BCDA yes
BDAC yes
BDCA no
CABD no
CADB yes
CBAD no
CBDA no
CDAB yes
CDBA yes
DABC yes
DACB no
DBAC no
DBCA no
DCAB yes
DCBA yes
I think that's it. 24, (4!), of those, but only 9 (those marked yes) have all four letters out of sequence.

So again, if I have counted correctly, 15*9=135. That's how many ways to have exactly two correct answers.

There are 720 arrangements possible and we have only accounted for 1+0+15+40+135=191 of them. There are 529 more arrangements. Some of those arrangements have one answer correctly placed and the rest have no answers correctly placed. If we find one, we can subtract from 529 to get the other.

I'll come back to this problem tomorrow. Maybe I'll think of something in the meanwhile. Almost sixty years ago as a junior in college I took a graduate math class for fun. I couldn't solve all the problems on the final exam. Several months after the class was finished, I was digging a ditch and not thinking about the problems on that final exam... and the solution to one of the problems popped into my head! Perhaps I had been subconsciously thinking about the problem all that time. Anyhow, maybe I'll subconsciously think about how to put either exactly one or exactly no answers in the correct place.

I don't think that works.

I agree.

I don't know what a cascading error is.

nickthegeek got the answers for 6, 5, 4, 3, and 2 correctly placed right. (At least we got the same answers). My guess is he also got 1 and 0 right... but I haven't figured out how to do either yet... maybe tomorrow.

It's a fun problem for me. Thanks.

Buzz

A=44

OK. I think I figured out a way to count one item (of six). Kind of tedious, but like I said, I’m not a mathematician.

Assuming 0,1,2,3,4,5 represent the six items, let 5 occupy the sixth position and then count all the possibilities using 0,1,2,3,4. Then go down the list (it will be 120 items long… like I say, kind of tedious… go down the list and identify all those that have none of the five in their proper place. (Marked “yes” in the second column below). However many yeses we have, we could do this six times.

01234	no
01243	no
01324	no
01342	no
01423	no
01432	no
02134	no
02143	no
02314	no
02341	no
02413	no
02431	no
03124	no
03142	no
03214	no
03241	no
03412	no
03421	no
04123	no
04132	no
04213	no
04231	no
04312	no
04321	no
10234	no
10243	no
10324	no
10342	yes
10423	yes
10432	no
12034	no
12043	yes
12304	no
12340	yes
12403	yes
12430	no
13024	no
13042	yes
13204	no
13240	no
13402	yes
13420	yes
14023	yes
14032	no
14203	no
14230	no
14302	yes
14320	yes
20134	no
20143	yes
20314	no
20341	yes
20413	yes
20431	no
21034	no
21043	no
21304	no
21340	no
21403	no
21430	no
23014	no
23041	yes
23104	no
23140	yes
23401	yes
23410	yes
24013	yes
24031	no
24103	yes
24130	no
24301	yes
24310	yes
30124	no
30142	yes
30214	no
30241	no
30412	yes
30421	yes
31024	no
31042	no
31204	no
31240	no
31402	no
31420	no
32014	no
32041	yes
32104	no
32140	yes
32401	yes
32410	yes
34012	yes
34021	yes
34102	yes
34120	yes
34201	no
34210	no
40123	yes
40132	no
40213	no
40231	no
40312	yes
40321	yes
41023	no
41032	no
41203	no
41230	no
41302	no
41320	no
42013	yes
42031	no
42103	yes
42130	no
42301	yes
42310	yes
43012	yes
43021	yes
43102	yes
43120	yes
43201	no
43210	no

There are 44 yeses (meaning for those 44, 5 is in the right place but all the others are not placed in their proper order.

Thus 44*6=264 is the number of ways we could have exactly one answer correct.

And then by inference,

• 720-1-0-15-40-135-264=265

265 is the number of way no answer is correct.

I see that kapw7 has also responded, but I have not looked at his solution yet. I’ll do that now.

Interesting problem.

A=44

Let N be the total number of objects (N=6 in OP).

Let X be the number of objects that are in the correct "slot" (so X can take values 0,1,2,...,N).

Let T(X;N) be the number of ways you can have X correct out of N objects.

Let C(A,B) be the number of ways you can choose B objects from A objects (C is the Combination)

Let D(Y) be the number of ways you can arrange Y objects in which 0 objects are in the correct slot (D is called the Derangement).

Then, it is clear that:

T(X;N) = C(N,X) * D(N-X)

since if there are X in the correct slots, N-X must be in incorrect slots. There are C(N,X) ways to choose the X out of N to be in correct slots. And D(N-X) ways for the N-X to be in incorrect slots.

There are several different ways to derive the D(Y) Derangement function. Principle of Inclusion-Exclusion is perhaps the most common.

I learned it via a simple recursive relationship. It is fairly straightforward to show that:

D(Y) = (Y-1) * [D(Y-1) + D(Y-2)] where D(0)=1 and D(1)=0.

----------

So, for OP's case of N=6 we have:

T(0;6) = C(6,0) * D(6)
T(1;6) = C(6,1) * D(5)
T(2;6) = C(6,2) * D(4)
T(3;6) = C(6,3) * D(3)
T(4;6) = C(6,4) * D(2)
T(5;6) = C(6,5) * D(1)
T(6;6) = C(6,6) * D(0)

which are easily evaluated.