448 Proceedings of Royal Society of Edinburgh. [skss. 
The mode of dealing with this converse problem will be under- 
stood from an example, say the case where 
8=2+3+ 0+1 +0. 
We look to the first item 2 and ask ourselves which of the five 
elements 1, 2, 3, 4, 5 must be put first so that there may be 
only two elements less than it to follow after, and the answer is 
clearly the element 3. Then 'we turn to the second item of 8, 
viz., 3, and ask which of the remaining elements 1, 2, 4, 5 must 
be put next so that there may be exactly three elements less than 
it to follow after, and the answer is the element 5. Again, we 
turn to the third item of 8, viz., 0, and ask which of the remaining 
elements 1, 2, 4 must be put next so that there may be no element 
less than it to follow after, and the answer is 1. Continuing this 
procedure we find the required permutation to be 
3 5 1 4 2. 
( 1 0) Taking the case of n = 4 and arranging the permutations in 
order, with the appropriate detailed value of 8 opposite each, we 
have the following table : — 
Ordinal No. 
Permutation. 
No. of Inverted-Pairs. 
1 
1234 
0 + 0 + 0 + 0 i. &. , 
0 
2 
1243 
0+0+1+0 
1 
3 
1324 
0+1+0+0 
1 
4 
1342 
0+1+1+0 
2 
5 
1423 
0+2+0+0 
2 
6 
1432 
0+2+1+0 
3 
7 
2134 
1+0+0+0 
1 
8 
2143 
1+0+1+0 
2 
9 
2314 
1+1+0+0 
2 
10 
2341 
l+l+l+O 
3 
11 
2413 
1+2+ 0+0 
3 
12 
2431 
1 + 2 + 1 + 0 
4 
13 
3124 
2+0+0+0 
2 
14 
3142 
2+0+1+0 
3 
15 
3214 
2+1 +0+0 
3 
16 
3241 
2+1+1+0 
4 
17 
3412 
2+2+0+0 
4 
18 
3421 
2+2+1 +0 
5 
19 
4123 
3+0+0+0 
3 
20 
4132 
3+0+1+0 
4 
21 
4213 
3+1+0+0 
4 
22 
4231 
3+1+1+0 
5 
23 
4312 
3+2+0+0 
5 
24 
i 
4321 
3+2+1+0 
6 
