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mathmath333
 one year ago
Sam has forgotten his friend’s sevendigit telephone number.
He remembers the following: the first three digits are either
635 or 674, the number is odd, and the number 9 appears
once. If Sam were to use a trial and error process to reach
his friend, what is the minimum number of trials he has to
make before he can be certain to succeed?
mathmath333
 one year ago
Sam has forgotten his friend’s sevendigit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number 9 appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{ Sam has forgotten his friend’s sevendigit telephone number.}\hspace{.33em}\\~\\ & \normalsize \text{ He remembers the following: the first three digits are either}\hspace{.33em}\\~\\ & \normalsize \text{ 635 or 674, the number is odd, and the number 9 appears }\hspace{.33em}\\~\\ & \normalsize \text{ once. If Sam were to use a trial and error process to reach }\hspace{.33em}\\~\\ & \normalsize \text{ his friend, what is the minimum number of trials he has to }\hspace{.33em}\\~\\ & \normalsize \text{ make before he can be certain to succeed?}\hspace{.33em}\\~\\ & a.)\ 1000 \hspace{.33em}\\~\\ & b.)\ 2430 \hspace{.33em}\\~\\ & c.)\ 3402 \hspace{.33em}\\~\\ & d.)\ 3006 \hspace{.33em}\\~\\ \end{align}}\)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1he had to make combinations like  1)number starting with635 dw:1441713928711:dw 1st we take the case where the number has 9 at the end and calculate possibilities nd then u take cases with 1,3,5,7 and fix 9 at one of these pointsdw:1441714010300:dw and calculate the possibilities nd then multiply it by 3 cause there are 3 such places :) so this way u get possibilities in this case :P similarly do it with 674 :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.110*9*9 for number ending with 1,3,5,7

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1and 9*9*9 for number ending with 9

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1nope :) 1) u have 4 numbers fixed like thisdw:1441714329392:dw so u have only 3 vacant places to be filled so u have total possibilities =9*9*4 and u can place the 9 at two other places too so u get total combinations=9*9*4*3

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1yea for 9 at the end it is 9*9*9 :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1but the difficulty is

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441714551371:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1no wait we haven't yet counted the case with674

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i m talkinh about this

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441714695772:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1:O yes i didn't think abt that so the number of combinations wuld be=9*9*10*3

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but the answer in book is 3402

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1so where did we go wrng :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1did the book allow repetitions

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1yes repetitions must be allowed if they were not then why wuld they specially say that 9 comes only once :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441715221286:dw hehe u haven't fixed the 9 so it can be like the 9 doen't even appears so this is wrong :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1what i did earlier ws correct u knw that 9 comes once nd lets say its not at the end so u have 3 places for it nd then u calculate like this dw:1441715357131:dw so u get 9*9*4 cases similarly u can move it to 3 alt places so u have a total of 9*9*4*3 combinations :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1omg imquerty is a genius again . O: :O

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1nd then if 9 is at the end u have 9*9*9 combinations total combinations =9*9*9 + 9*9*4*3 =1701 and u multiply it by 2 cause u also get same number of combinations for 674 so answer is 1701x2 =3402

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1do u have querty keypad

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i m angry on myself , i cant ssolve counting questions

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1lol i've learnt so much permutation and combination questions haha we haven't started it yet :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1u jst don need to do anything except multiply numbers nd think ways ways to adjust :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1ahahah looks like hes goin to tumble :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1monks dont tumble, sumos tumble

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Hi, This is what I have.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1hi, i like it what u have :)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Considering only the last four digits Pattern(XXX9)=9^3=729 ways Patterns(9XX, X9X, XX9 {1,3,5,7})=3*4*9^2=972 Total for last 4 digits = 729+972=1701 Multiply by two prefixes, 2*1701=3402
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