mathmath333
  • mathmath333
Sam has forgotten his friend’s seven-digit 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ Sam has forgotten his friend’s seven-digit telephone number.}\hspace{.33em}\\~\\ & \normalsize \text{ He remembers the following: the first three digits are either}\hspace{.33em}\\~\\ & \normalsize \text{ 635 or 674, the number is odd, and the number 9 appears }\hspace{.33em}\\~\\ & \normalsize \text{ once. If Sam were to use a trial and error process to reach }\hspace{.33em}\\~\\ & \normalsize \text{ his friend, what is the minimum number of trials he has to }\hspace{.33em}\\~\\ & \normalsize \text{ make before he can be certain to succeed?}\hspace{.33em}\\~\\ & a.)\ 1000 \hspace{.33em}\\~\\ & b.)\ 2430 \hspace{.33em}\\~\\ & c.)\ 3402 \hspace{.33em}\\~\\ & d.)\ 3006 \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
r ut yping
imqwerty
  • imqwerty
he 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 points-|dw: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 :)

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More answers

mathmath333
  • mathmath333
10*9*9 for number ending with 1,3,5,7
mathmath333
  • mathmath333
and 9*9*9 for number ending with 9
imqwerty
  • imqwerty
nope :) 1) u have 4 numbers fixed like this-|dw: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
  • imqwerty
yea for 9 at the end it is 9*9*9 :)
imqwerty
  • imqwerty
but the difficulty is
imqwerty
  • imqwerty
the options :)
mathmath333
  • mathmath333
|dw:1441714551371:dw|
imqwerty
  • imqwerty
no wait we haven't yet counted the case with674
mathmath333
  • mathmath333
see my photu
imqwerty
  • imqwerty
yea LOL :) fat monk
mathmath333
  • mathmath333
i m talkinh about this
mathmath333
  • mathmath333
|dw:1441714695772:dw|
imqwerty
  • imqwerty
:O yes i didn't think abt that so the number of combinations wuld be=9*9*10*3
mathmath333
  • mathmath333
but the answer in book is 3402
imqwerty
  • imqwerty
so where did we go wrng :)
mathmath333
  • mathmath333
did the book allow repetitions
imqwerty
  • imqwerty
hey wait
imqwerty
  • imqwerty
no nvm
imqwerty
  • imqwerty
yes repetitions must be allowed if they were not then why wuld they specially say that 9 comes only once :)
mathmath333
  • mathmath333
ok
imqwerty
  • imqwerty
hey wait
imqwerty
  • imqwerty
|dw: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
  • imqwerty
what 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
  • mathmath333
omg imquerty is a genius again . O: :O
imqwerty
  • imqwerty
nd 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
imqwerty
  • imqwerty
lol
mathmath333
  • mathmath333
do u have querty keypad
imqwerty
  • imqwerty
on my laptop :) yea
mathmath333
  • mathmath333
i m angry on myself , i cant ssolve counting questions
imqwerty
  • imqwerty
lol i've learnt so much permutation and combination questions haha we haven't started it yet :)
imqwerty
  • imqwerty
u jst don need to do anything except multiply numbers nd think ways ways to adjust :)
imqwerty
  • imqwerty
ahahah looks like hes goin to tumble :)
mathmath333
  • mathmath333
monks dont tumble, sumos tumble
imqwerty
  • imqwerty
:O
imqwerty
  • imqwerty
༼•͟ ͜ •༽
mathmate
  • mathmate
Hi, This is what I have.
imqwerty
  • imqwerty
hi, i like it what u have :)
mathmate
  • mathmate
Considering 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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