This name has been given to a contrivance said to have originated with the famous Pascal, or to have been perfected by him.
1
2 1
3 3 1
4 6 4 1
5 10 10 5 1
6 15 20 15 6 1
7 21 35 35 21 7 1
8 28 56 70 56 28 8 1
&c. &c.
This peculiar series of numbers is thus formed: Write down the numbers 1, 2, 3, &c., as far as you please, in a vertical row. On the right hand of 2 place 1, add them together, and place 3 under the 1; then 3 added to 3 = 6, which place under the 3; 4 and 6 are 10, which place under the 6, and so on as far as you wish. This is the second vertical row, and the third is formed from the second in a similar way.
This triangle has the property of informing us, without the trouble of calculation, how many combinations can be made, taking any number at a time out of a larger number.
Suppose the question were that just given; how many selections can be made of 3 at a time out of 8? On the horizontal row commencing with 8, look for the third number; this is 56, which is the answer.
HOW MANY DIFFERENT DEALS CAN BE MADE WITH 13 CARDS OUT OF 52?
To discover this we must make a continued multiplication of 52 × 51 × 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 × 40, being 13 terms for the 13 cards, also a continued multiplication of 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, and having found the two products, we must divide one by the other, and the quotient is the number of different deals out of 52 cards. This "sum," that looks so formidable with natural figures, is a very short one by logarithms.
THE THREE GRACES.
Three articles, or three names inscribed on cards, having been distributed between three persons, you are to tell which article or card each person has.
Designate the three persons in your own mind, as 1st, 2d, and 3d, and the three articles, A, E, I. Provide 24 counters, and give 1 to the first person, 2 to the 2d, 3 to the 3d. Place the remaining 18 on the table. Request that the three persons will distribute among themselves the three articles, and that, having done so, the person who has the one which you have secretly denoted by A, will take as many counters as he may have already; the holder of E must take twice as many as he may have; and the holder of I must take four times as many. Then leave the room, in order that the distribution of articles and of counters may be made unobserved by you. We will suppose that the three articles are three cards, on which are the words Clara, Rosa, Emily, which you will yourself secretly denote by the letters A, E, I. Suppose also that in the division the first person has Emily (I), the second has Clara (A), and the third has Rosa (E), then the 1st will take four times as many counters as he has (1), and will therefore take 4; the 2d will take as many as he has (2), and will therefore take 2; the 3d will take 6, being twice as many as he has (3). On the table will be left 6 counters. The distribution having been made, you will return and observe the number of counters on the table, from which you can find who is the holder of each card by the following method.