PROBLEM 1. To find the number of permutations, or changes, that can be made of any given number of things, all different from each other. RULE.* Multiply all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product will be the answer required. EXAMPLES. 1. How many changes may be made with these three letters, abc? 1 2 Or 1×2×3=6 the answer. 2 3 የ * The reason of the rule may be shown thus: any one thing is capable only of one position, as a. Any two things, a and b, are only capable of two variations; as ab, ba; whose number is expressed by 1×2. If there be 3 things, a, b, and c, then any two of them, leaving out the third, will have 1×2 variations; and consequently, when the third is taken in, there will be 1×2×3 variations. In the same manner, when there are 4 things, every 3, leaving out the forth, will have 1×2×3 variations. Then, the fourth being taken in, there will be 1×2×3×4 variations. And so on, as far as you please. the changes. abc acb bac bca cab cba 2. How many changes may be rung on 6 bells? Ans. 720. 3. For how many days can 7 persons be placed in a different position at dinner? Ans. 5040 days. 4. How many changes may be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in 1 minute, and the year to consist of 365 days 5 hours and 49 minutes? Ans. 479001600 changes, and 91y. 26d. 22h. 41m. 5. How many changes may be made of the words in the following verse? Tot tibi sunt dotes, virgo, quot sydera cœlo. Ans, 40320. PROBLEM 2. Any number of different things being given, to find how many changes can be made out of them, by taking any given number at a time. RULE.* Take a series of numbers, beginning at the number of things given, and decreasing by 1 till the number of terms * This rule, expressed in terms, is as follows: mxm- X m2×m—3, &c. to n terms; where m = number of thing given, and n = quantities to be taken at a time. be equal to the number of things to be taken at a time, and the product of all the terms will be the answer required. In order to demonstrate the rule, it will be necessary to premise the following LEMMA. The number of changes of m things, taken n at a time, is equal to m changes of m-1 things, taken n-1 at a time. DEMONSTRATION. Let any 5 quantities, abcde, be given. First, leave out the a, and let v = number of all the variations of every two, bc, bd, &c. that can be taken out of the 4 remaining quantities, bcde. Now let a be put in the first place of each of them, abc, abd, &c. and the number of changes will still remain the same; that is, v = number of variatons of every 3 out of the 5, abcde, when a is first. In like manner, if b, c, d, e, be successively left out, the number of variations of all the twos will also =v; and b, c, d, e, ber ing respectively put in the first place, to make 3 quantities out of 5, there will still be v variations as before. But these are all the variations, that can happen of 3 things out of 5, when a, b, c, d, e, are successively put first; and therefore the sum of all these is the sum of all the changes of 3 things out of 5. But the sum of these is so many times v, as is the number of things; that is, 5v, or mv, all the changes of three things out of 5. And the same way of reasoning may be applied to any numbers whatever. DEMONSTRATION OF THE RULE. Let any 7 things, abcdefg, be given, and let 3 be the number of quantities to be taken. Then m7, and n=3. EXAMPLES. 1. How many changes may be made out of the 3 letters, abc, by taking 2 at a time. alphabet, it being admitted, that a number of consonants may words can be made with 5 letters of the make a word ?. PROBLEM 3. Ans. 5100480. Any number of things being given, whereof there are several given things of one sort, several of another, &c. to find how many changes can be made out of them all. Now it is evident, that the number of changes, that can be made by taking 1 by 1 out of 5 things, will be 5, which let =v. Then, by the lemma, when m=6 and n=2, the number of changes will =mv=6×5; which let v a second time. Again by lemma, when m=7 and n=3, the number of chang es=mv=7×6×5; that is,m v=mxm—1xm—2, continued to 3, or n terms. And the same may be shown for any other num bers. RULE.* 1. Take the series 1, 2, 3, 4, &c. up to the number of things given, and find the product of all the terms. * This rule is expressed in terms thus: 1X2X3X4×5, &c. to m 1x2x3, &c. to of things given, p X1×2×3, &c. to q, &c. ber of things of the ; where m number number of things of the first sort, 9 = numsecond sort, &c. The DEMONSTRATION may be shown as follows. Any two quantities, a, b, both different, admit of 2 changes; but if the quantities be the same, or ab become aa, there will be but one alteration, which may be expressed by 1x2 1x2 1. Any three quantities, abc, all different from each other, afford 6 variations; but if the quantities be all alike, or abc become aaa, then the 6 variations will be reduced to 1, which may be I×2×3 1×2×3 expressed by 1. Again, if two of the quantities on ly be alike, or abc become aac, then the six variations will be reduced to these 3, aac, caa, and aca, which may be expressed by 1x2x3 1X2 = 3. Any four quantities, abcd, all different from each other, will admit of 24 variations; but if the quantities be the same, or abcd become aaaa, the number of variations will be reduced to 1×2×3×4 one; which is = 1×2×3×4 1. Again, if three of the quan tities only be the same, or abcd become aaab, the number of variations will be reduced to these 4, aaab, aaba, abaa, and baaa, 1×2×3×4 which is 1X2X3 4. And thus it may be shown, that, |