Quiz 3 Sol. - 2602Q3Ansnb

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2602Q3Ansnb http://www.math.gatech.edul~morley/2602/quizes/spring02/Q3/index.htrnl Math 2602 Math 2602 Feb 14 Problem 1 (10 points) a) Jenny Thompson would like you to find out how many ways there are to put 7 indistinguishable balls into 4 indistinguishable urn with NO EMPTY urns. Just write out all the ways. Use this result to answer the same question with distinguishable Balls. Put 1 into each urn. There are 3 more to be distributed. The ways to to this are: u 2 1, 1 and , 2, 2, 1 So there are three ways to do this. If the balls are distinguishable then 4, 1, 1, 1 contributes [Z], 3, 2, 1, 1 contributes and 2, 2, 2, 1 contributres (this is the tricky one) / 3! Add 'em up: Ih[148]:= Binomialfi, 4] + Binmialfi, 3] Bjnmnia1[4, 2] + Bjnm-nialfi, 2] Binmia1[5, 2] Binmnia1[3, 2] f3: 0ut[143]= 35!] Problem 2 (10 Points) Dara Tores ywould like you to find the series for (1 + 27) 1’3 . Explicitly find the first 3 or 4 terms. 'I'hi3isju3tlike any)“, x = 1, y: 27 Binomialseriea. °° 1 3 Z[ 1: ]1k Hug—k, whichcanbe rewritten in avarietyofways. k=D M149]: term[k_] := Bjnm'nialflfl lt]2?“1m"‘J Here‘ 3 awhole bunch" in terms. 1 of 3 09/03/2003 11:46 PM 2602Q3.A_ns.nb ht ://www.math.gatech.edu/~morley/2602/quizes/spring02/Q3/index.hlml tP ln[151]:= Table[{k, tem[k]}, {k, I], 10}] H TableFum o 3 1 1 37 1 2 "m7 5 3 531441 10 4 _43046721 22 5 3436734401 154 6 _E4?283609443 374 7 63630377364333 335 8 '5559060566555523 21505 9 4052555153013376267 55913 10 '323256967394537077627 Problem 3 (10 points) Janet Evans would like you to count the number of numbers from 0000 to 9999 that do not contain 2 ( or more) adjacent 6's (like 5664 or 6631) or two (or more) adjacent 7's. Numbers With (exaxtly) 2 adjacent 6's 66NX N66N XN 66 N = anthing but a six. 9 possibilities X = anything 10 possibilities. The total so far is 90 for 66NX, 81 f0 X66X, and 90 for XN66 Three 6;s 666N or N666 9 + 9 ways Four 6's 6666 1 way. Total ln[152]:= 90+ 81 + 9|] + 18 + 1 2 of 3 09/03/2003 11:46 PM 2502Q3Ar s.nb _ http://www.math.gatech.edu/~mor1ey/2602/quizes/sp1ing02/Q3/index.html 0ut[152]= 280 Same for 7s' 280 bad ones. There is 2 ways to get adjacent 6's and 7's: 6677 ort 7766. Putting this all together we get (for the answer) |n[153]:= lflflflll — 2280 + 2 0ut[153]= 9442 Ans Converted by Mathematica February 21, 2002 3 of 3 09/03/2003 11:46 PM ...
View Full Document

This note was uploaded on 05/12/2010 for the course MATH 2602 taught by Professor Costello during the Spring '09 term at Georgia Tech.

Page1 / 3

Quiz 3 Sol. - 2602Q3Ansnb

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online