awais Posted November 8, 2004 Share Posted November 8, 2004 Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7? Q.2. A five digit number is chosen at random, What is the probability that the product of the digits is an even number? Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 8, 2004 Share Posted November 8, 2004 Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7? Q.2. A five digit number is chosen at random, What is the probability that the product of the digits is an even number? Q1. 7*7 *4 = 49*4 = 196 (if no repetition, then 6*5 *4 =30*4 = 120) Q2. 6561*9 729*7 81*5 9*3 Total=64584 Total 5digit numbers=90000 P = 64584/90000 ~ 0.71 Quote Link to comment Share on other sites More sharing options...
DMJ Posted November 8, 2004 Share Posted November 8, 2004 Q1 a) with repetition 7*7*4=196 b)without repetition first two digits -both even 3!/1!=6 times number of odd digits left *4 =24 first two digits -both odd 4!/2!=12 times number of odd digits left *2 =24 first two digits -one even one odd 3*4=12 times number of odd digits left *3=36 asw =84 Q2. in order to have product of any number of numbers even, one of numbers must be even. The probability that all five digits are odd (assuming that 0 is even) is: (5/10)^5 = (1/2)^5 =1/32 the probability that at least one of digits is even =31/32 asw 31/32 Quote Link to comment Share on other sites More sharing options...
shaq Posted November 8, 2004 Share Posted November 8, 2004 Q#1 1)with repetition 7 * 7* 4 = 196 2)without repetition a) odd odd odd = 4*3*2 = 24 b) odd even odd = 4* 3* 3 = 36 c) even odd odd = 3 * 4* 3 = 36 d) even even odd = 3* 2* 4= 24 Total 120 Q#2 1- ( probability that all numbers are odd) 1 - ( (1/2)^5) = 31/32 Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 8, 2004 Share Posted November 8, 2004 yap. second is 31/32 just learned th method segregation from another thread. Quote Link to comment Share on other sites More sharing options...
shaq Posted November 8, 2004 Share Posted November 8, 2004 Q1. (if no repetition, then 6*5 *4 =30*4 = 120) Hey bull, how did you get this ? Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 8, 2004 Share Posted November 8, 2004 Hey bull, how did you get this ? :) u testin' me or u din't actually get it... gotta be kiddin me, i learned P&C & P analysin ur answers & studyin links provided by u.. Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7? odd no.s-- 1,3,5,7 Case 1-- Last digit is 1 (fixed) so other positions can be filled in (7-1=6): 6*5 ways similarly for other 3 odd numbers. Hence, 6*5 *4=120 how did i do???;) Quote Link to comment Share on other sites More sharing options...
awais Posted November 8, 2004 Author Share Posted November 8, 2004 Answer for the first question, 7*7*4= 196 For the second, 1- ( 5/10 )^5 or 1 - (1/2)^5 = Hey Bullfighter good job for solving q.1 without repetition. Quote Link to comment Share on other sites More sharing options...
shaq Posted November 8, 2004 Share Posted November 8, 2004 :) u testin' me or u din't actually get it... gotta be kiddin me, i learned P&C & P analysin ur answers & studyin links provided by u.. Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7? odd no.s-- 1,3,5,7 Case 1-- Last digit is 1 (fixed) so other positions can be filled in (7-1=6): 6*5 ways similarly for other 3 odd numbers. Hence, 6*5 *4=120 how did i do???;) Good job bull. I split it up differently ( but yours was simpler). Kudos to you kid..:tup: Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 8, 2004 Share Posted November 8, 2004 heeey... thanx man. (shaq just praised mine over 'shaq's' own method, tht too for P.. now, this is an honour.) :grad: Quote Link to comment Share on other sites More sharing options...
awhig Posted November 8, 2004 Share Posted November 8, 2004 Q.1. How many three digit odd numbers can be formed using the digits 1,2,3,4,5,6 and 7? Q.2. A five digit number is chosen at random, What is the probability that the product of the digits is an even number? In the first question what should be assumed about the repetion of digits by default??????? Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 8, 2004 Share Posted November 8, 2004 In the first question what should be assumed about the repetion of digits by default??????? As nothing is mentioned, and also as the word 'using' is used--- repetitions r allowed. if Q was, from given digits 1,2,3,3,4,5,5,5,6,9 or other wise mentioned then repetitions not allowed. would be mentioned. Quote Link to comment Share on other sites More sharing options...
awhig Posted November 8, 2004 Share Posted November 8, 2004 As nothing is mentioned, and also as the word 'using' is used--- repetitions r allowed. if Q was, from given digits 1,2,3,3,4,5,5,5,6,9 or other wise mentioned then repetitions not allowed. would be mentioned. Thanks. Quote Link to comment Share on other sites More sharing options...
Malegria Posted November 9, 2004 Share Posted November 9, 2004 Please to understand more this kind of problems.. where I can find P&C theory?? thanks Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 9, 2004 Share Posted November 9, 2004 http://regentsprep.org/regents/math/math-topic.cfm?TopicCode=permut http://regentsprep.org/regents/math/math-topic.cfm?TopicCode=combin Quote Link to comment Share on other sites More sharing options...
BuZZ Posted November 9, 2004 Share Posted November 9, 2004 Q.2. A five digit number is chosen at random, What is the probability that the product of the digits is an even number? All possibilities of 5 digit numbers such that all digits are odd = 5^5 All possibilities of 5 digit number = 9*(10)^4 Required Ans = 1 - ([5^5]/[9*(10)^4]) = 139/144 Quote Link to comment Share on other sites More sharing options...
Malegria Posted November 9, 2004 Share Posted November 9, 2004 thebullfighter ? why 1 is fixed?? in your answer you did not start with 7? Thanks .. Quote Link to comment Share on other sites More sharing options...
thebullfighter Posted November 9, 2004 Share Posted November 9, 2004 thebullfighter ? why 1 is fixed?? in your answer you did not start with 7? Thanks .. dear for Odd number, The last Digit also has to be Odd. Now, we have 4 Odd numbers... (don't remember wat th exact Q was.) so, for 4Digit number to be odd, The last Place/Position/Digit can take Only Four values... For, other 3places as explained possiblities are as normally counted. (so, we fix or freeze the last Digit/Position, and solve th Q as for a 3Digit number as normal... hence, hope it's clear now.. else let know.. will take up in detail Quote Link to comment Share on other sites More sharing options...
Malegria Posted November 9, 2004 Share Posted November 9, 2004 Thanks I realized that :) Quote Link to comment Share on other sites More sharing options...
teenaja Posted November 14, 2005 Share Posted November 14, 2005 i agree with buzz. Because 0 cannot start so that All possibilities of 5 digit number = 9*10*10*10*10 Quote Link to comment Share on other sites More sharing options...
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