In the first edition, vol i. p. 158, I stated that I had not been able to discover any mode of calculating the number of cases in which inconsistency would be implied in the selection of combinations from the Logical Alphabet. The logical complexity of the problem appeared to be so great that the ordinary modes of calculating numbers of combinations failed, in my opinion, to give any aid, and exhaustive examination of the combinations in detail seemed to be the only method applicable. This opinion, however, was mistaken, for both Mr. R. B. Hayward, of Harrow, and Mr. W. H. Brewer have calculated the numbers of inconsistent cases both for three and for four terms, without much difficulty. In the case of four terms they find that there are 1761 inconsistent selections and 63,774 consistent, which with one case where no condition exists, make up the total of 65,536 possible selections. The inconsistent cases are distributed in the manner shown in the following table:-- +--------------+---------------------------------------------------+ | Number of | | | Combinations | 0 1 2 3 4 5 6 7 8 9 10, &c. | | remaining. | | +--------------+---------------------------------------------------+ | Number of | | | Inconsistent | 1 16 112 352 536 448 224 64 8 0 0, &c. | | Cases. | | +--------------+---------------------------------------------------+ When more than eight combinations of the Logical Alphabet (p. 94, column V.) remain unexcluded, there cannot be inconsistency. The whole numbers of ways of selecting 0, 1, 2, &c., combinations out of 16 are given in the 17th line of the Arithmetical Triangle given further on in the Chapter on Combinations and Permutations, the sum of the numbers in that line being 65,536. *Professor Clifford on the Types of Compound Statement involving Four Classes.*