Math 141, Fall 2009,cBenjamin AurispaExtra Counting Practice Solutions1. How many 6-digit numbers are possible if the first two digits cannot be 0, and there can be norepeated digits?1st digit: Cannot be 0 – 9 choices2nd digit: Cannot be 0 or the first digit – 8 choices3rd digit: Cannot be first or second digit – 8 choices4th digit: Cannot be first, second, or third digit – 7 choices5th digit: Cannot be first, second, third, or fourth digit – 6 choices6th digit: Cannot be first, second, third, fourth, or fifth digit – 5 choices9 · 8 · 8 · 7 · 6 · 5 = 120,9602. How many 5-digit numbers are possible if the first digit must be odd, the second digit cannot be 0,and the first 3 digits must be different?1st digit: Must be odd – 5 choices2nd digit: Cannot be 0 or the first digit – 8 choices3rd digit: Cannot be first or second digit – 8 choices4th digit: No restrictions – 10 choices5th digit: No restrictions – 10 choices5 · 8 · 8 · 10 · 10 = 32,0003. How many 6-digit numbers are possible if the last digit must be even but not 0, the first digitcannot be 0 or 1 and there can be no repeated digits?Start with choosing last digit, since it has the most restrictions.6th (last) digit: Must be even, not 0 – 4 choices1st digit: Cannot be 0, 1, or last digit – 7 choices2nd digit: Cannot be first or last digit – 8 choices3rd digit: Cannot be first, second, or last digit – 7 choices4th digit: Cannot be first, second, third, or last digit – 6 choices5th digit: Cannot be first, second, third, fourth, or last digit – 5 choices7 · 8 · 7 · 6 · 5 · 4 = 47,0404. How many 5-letter “words” are possible if the word must start and end with a vowel and theremaining letters must be different consonants?1st letter: Must be a vowel – 5 choices5th (last) letter: Must be a vowel – 5 choices2nd letter: Must be a consonant (not a vowel) – 21 choices3rd letter: Must be a consonant, cannot be 2nd letter – 20 choices4th letter: Must be a consonant, cannot be 2nd or 3rd letter – 19 choices5 · 21 · 20 · 19 · 5 = 199,5005. A jar contains 21 jelly beans: 4 red, 5 green, 7 white, 3 black, and 2 orange. A sample of 6 jellybeans is chosen.(a) How many samples are possible?C(21, 6) = 54,2641Math 141, Fall 2009,cBenjamin Aurispa(b) How many samples contain no whites?Sample contains: 0 white, 6 not whiteC(7, 0)C(14, 6) = C(14, 6) = 3,003(c) How many samples contain exactly 2 black?Sample contains 2 black, 4 not blackC(3, 2)C(18, 4) = 9,180(d) How many samples contain exactly 4 white and exactly 2 red?C(7, 4)C(4, 2) = 210(e) How many samples contain exactly 3 green and exactly 2 black?Sample contains 3 green, 2 black, 1 not green or blackC(5, 3)C(3, 2)C(13, 1) = 390(f) How many samples contain exactly 3 red or exactly 5 white?Two cases combined using union rule:n(ex 3 R or ex 5 W)= n(ex 3 R)+n(ex 5 W)−n(ex 3 R and ex 5 W)n(ex 3 R and ex 5 W)= 0 since it is not possible to have 8 jelly beans in a sample of 6.3 red, 3 not red 5 white, 1 not whiteC(4, 3)C(17, 3) + C(7, 5)C(14, 1) = 3,014(g) How many samples contain exactly 2 black or exactly 4 green?Two cases combined using union rule:n(ex 2 B or ex 4 G)= n(ex 2 B)+n(ex 4 G)−n(ex 2 B and ex 4 G)2 B, 4 not B 4 G, 2 not G 2 B, 4 GC(3, 2)C(18, 4) + C(5, 4)C(16, 2) − C(3, 2)C(5, 4) = 9,765(h) How many samples contain at least 1 white?Total number of samples − Number of samples with no whiteC(21, 6) − C(14, 6) = 51,261(i) How many samples contain at least 3 red?At least 3 red means exactly 3 red OR exactly 4 red. (There can’t be more than 4 red.)C(4, 3)C(17, 3) + C(4, 4)C(17, 2) = 2,856(j) How many samples contain at most 3 green?At most 3 green means 0, 1, 2, OR 3 green.One method is to add up these four cases.C(5, 0)C(16, 6) + C(5, 1)C(16, 5) + C(5, 2)C(16, 4) + C(5, 3)C(16, 3) = 53,648Another method is to take the total number of samples and subtract the number of samplesyou don’t want (the complement). The samples you don’t want are those with 4 or 5 green.C(21, 6) − [C(5, 4)C(16, 2) + C(5, 5)C(16, 1)] = 53,6482Math 141, Fall 2009,cBenjamin Aurispa6. John has 12 CD’s in his car: 3 are country, 3 are rap, 4 are classical, and 2 are blues.(a) In how many ways can all 12 CDs be arranged in a CD holder?12! = 479,001,600(b) In how many ways can 9 of 12 CDs be arranged in a CD holder?P (12, 9) = 12 · 11 · . . . · 4 = 79,833,600(c) In how many ways can all 12 CD’s be arranged in a CD holder if each type of music is puttogether?Arrange the four groups (types) of music: 4!Arrange CDs in country group: 3!Arrange CDs in rap group: 3!Arrange CDs in classical group: 4!Arrange CDs in blues group: 2!4! · 3! · 3! · 4! · 2! = 41,4727. In how many distinguishable ways can the letters in the word SENSELESSNESS be arranged?Total number of letters: 13Number of identical S’s: 6Number of identical E’s: 4Number of identical N’s: 2Number of identical L’s: 113!6! · 4! · 2!= 180,1808. A boss is interviewing candidates in order to hire a manager, 2 computer programmers, 3 programcoordinators, and 4 secretaries. The secretaries will be chosen from a pool of 9 candidates and theremaining positions will be chosen from among 8 qualified candidates. In how many ways can theboss fill these positions?Number of ways to hire manager: C(8, 1) = 8Number of ways to hire 2 computer programmers: C(7, 2)Number of ways to hire 3 program coordinators: C(5, 3)Number of ways to hire 4 secretaries: C(9, 4)8 · C(7, 2) · C(5, 3) · C(9, 4) =
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