Question

Question 1 (a) How many positive integers are there between 1000 and 4999, inclusive? (b) How many positive integers between 1000 and 4999, inclusive: 1. have no repeated digit? 2. have at least one repeated digit? 3. have at most two repeated digits? Note that by 'one repeated digit' we mean that there is a digit that appears at least twice (eg, 1123 has one repeated digit). Similarly, by two repeated digits we mean a digit that appears at least three times (eg, 1112). Question 2 Provide the formulas. You do not need to evaluate these formulas, although you may if you wish. We throw a die six times. For each throw you record the outcome: EVEN (you throw a two, four or six) or ODD (you throw a one, three or five). Calculate the following 1. How many possible outcomes are there? 2. How many possible outcomes contain exactly 3 EVEN throws? 3. How many possible outcomes contain at most 1 ODD throw?

Answer 1

QO@The number of possitive fstegers between 1000 and 4999 including them are = 4999 - 1000 + 1 = 3999 + 1 - 4000] So, the sample space , 151 e you DNo repeate deget We will use counting principle hore, As we can use only 1.2.3.4 in the firest place as all the numbers are from 1000 to yaaa. So it can be done in uc, = 4 ways. 49 We can use any number from 0,1,2,3,4,5, 6, 7, 8, 9 fore the second place . But as there shouldn't be any repeatation of digits. So we have to skip the number that is selected for the finest place and choose from the rest nine digets and this can be done in ar-aways. The same method is used for choosing the to number for the third place by leaving numberes from 1st and 2nd place and this can be done in Sci-8 ways. procedure was followed fore the forth place, same done in tc 7 ways. and

So, the total number of ways we can choose the positive integere from 1000 and uggg that have no repeated digers = 4 x9x8x7 2016 ways At least one repeated digit Total number of ways we can choose a number with at least one repeated digit - (Total numbers) - (Numbers having no digits - repeated - 4000 - 2016 =liasy ways. ) tin At most two repeated digits - Numbers haring all digets the same ore z digers repeated arie 1111, 2222, 3335, 4444 so there are in total 4 numbers. : Numbers having at most two repeated digats - Total numberes) - (repeated -Numberes having & digets = 4000 - 4 3996 numbers

(@2) ( if we throw a times we throw die six times, then et we can have six each outcomes, So, the total number of outcomes are - 6 X 6 X 6 X6 X6 X6 - 66 - 279,936 . (1) We can get an EVEN throw when the outcome is either 2, 4 or 6. So it can be done in 3 ways. Exactly 5 Even throws means we have 3 EVEN throws and 3 ODD throws. - FLETE] 010] ol 3 3 3 3 3 3 (EVED) (ODD) 3 EVEN Throws can be done in = 3x3x3 - 27 ways. Similarly, 3 ODD Horows can be done to 3x3x3 - 27 ways

ろ ele Polóð So, this can be done のろべろべろ×ろメスメス -36=729 ways. Now, these even and odd throws can be Sh rearranged is G! ways. Z! 3! So, the total number of ways in which coe con get exactly 3 EVEN throws O x 3 x 36 31X2 X 36 COM) Possible outcomes containing No ODD outcomes ont all oven outcomes - [ETEE EEE 3 3 3 3 3 3 86 ways. Possible outcomes having one ODD parent throw - 6 X 36 as we can choose one ODD throw and rest even throws in 36 ways and the ODD throw can be suffled in 6 ways. 50, possible outcomes containing at most | ODD thrrow = (NO ODD) & (one ODD' (throw throw 26+ (6x36) - 38(6+1) =(7 x 36 )ways