Therefore, when there are no repetitions in the combinations, we have this RULE. Add one to the number. of individual things, and from the sum subtract the number denoting the class of the combinations, multiply this remainder by the successive increasing terms of the series of natural numbers, till we reach the term denoting the number of individual things ; then divide this product by the number of permutations, found by Case I., Art.88, of a number of individual things, denoted by the class of the combina. tions. Examples. 1. What is the number of combinations, without repetitions, · of the fourth class, which can be formed out of 10 individual things ? :: BOLUTION. Adding one to the number of individual things, we get 11, from which, subtracting the number denoting the class, we fon 7X8X9X10 get 11-437. Therefore, =210, is the num s 1x2x3x4 ber sought. 2. How many lottery tickets, each having three numbers, can be formed out of 60 numbers ? s. 58 x 59 x 60_34220. 3. How many combinations, of 4 things at a time, without repetitions, can be formed out of 100 individual things ? VARIATION. 90. By VARIATION, we understand the different transpo. sitions that can take place when the individual things are not all taken at once. Variations are divided into classes, in the same way as Combinations. They are also distinguished by variations with repetitions, and variations without repetitione. CASE I. When the variations are made with repetitions, we have as follows: . VARIATIONS WITH REPETITIONS. : 1) Var. (a, b, c, d, e, f) . (Second Class.) ices berec aaaa ici ara coca CITURE caub mic ccan eche babc beta RULE. Raise the number denoling the individual things to a power whose exponent is the number expressing the class of variations. Examples. . . . · 1. How many variations, with repetitions, can be formed of the 4th class, out of 5 individual things ? Ans. 54=625. 2. How many numbers can we form out of the nine dig. gits, of 9 places of figures each, provided we are allowed to make a repetition of the digits ? . Ans. 9'=387420489. CASE II. When the variations are formed, without repetitions, we find as follows: VARIATIONS WITHOUT REPETITIONS. Var. (a, b, c, d, e) abcd aecb bdea cdab dbca eadb bdece cdae dbce . eadc dcae . - deba ebda acbe bade bedc ebdc cabd ceba dcea "dabc deba ecda eabc edba adec bdae code daec eabd aebc, bdca cbea d bac eacb edca aebd bdce cbed , dbae eacd edcb From the above, we deduce the following RULE. Add one to the number of individual things, and from the sum subtract the number denoting the order of the class of va. riations ; multiply this remainder continually by the successive increasing terms of the series of natural numbers, until we reach that term which is equal to the number of individual things. . . Examples. .. . 1. How many variations, without repetitions, of the 4th class, can be formed out of 10 individual things ? Ans. 7X8X9x10=5040. 2. How many numbers, of five places of figures each, can be formed out of the 9 digits, so that the same digit shall not be repeated in the same number? Ans. 5 X 6 X7X8X9=15120. 3. How many words, of 13. letters each--no letter being repeated—can be formed out of the 26 letters of the alpha. bet, on the supposition that every combination of letters is capable of producing a word ? 91. It is required to find the number of ways in which an odd number of letters, arranged as in the following squares, can be read in alphabetical order, by beginning at the central square, and reading outwards. - 1 clbic By carefully inspecting the above figures, we derive the following |