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?

- mathmath333

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

r ut yping

- 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 :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- mathmath333

10*9*9 for number ending with 1,3,5,7

- mathmath333

and 9*9*9 for number ending with 9

- 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

yea for 9 at the end it is 9*9*9 :)

- imqwerty

but the difficulty is

- imqwerty

the options :)

- mathmath333

|dw:1441714551371:dw|

- imqwerty

no wait we haven't yet counted the case with674

- mathmath333

see my photu

- imqwerty

yea LOL :) fat monk

- mathmath333

i m talkinh about this

- mathmath333

|dw:1441714695772:dw|

- imqwerty

:O yes i didn't think abt that so the number of combinations wuld be=9*9*10*3

- mathmath333

but the answer in book is 3402

- imqwerty

so where did we go wrng :)

- mathmath333

did the book allow repetitions

- imqwerty

hey wait

- imqwerty

no nvm

- imqwerty

yes repetitions must be allowed if they were not then why wuld they specially say that 9 comes only once :)

- mathmath333

ok

- imqwerty

hey wait

- 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

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

omg imquerty is a genius again . O: :O

- 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

lol

- mathmath333

do u have querty keypad

- imqwerty

on my laptop :) yea

- mathmath333

i m angry on myself , i cant ssolve counting questions

- imqwerty

lol i've learnt so much permutation and combination questions haha we haven't started it yet :)

- imqwerty

u jst don need to do anything except multiply numbers nd think ways ways to adjust :)

- imqwerty

ahahah looks like hes goin to tumble :)

- mathmath333

monks dont tumble, sumos tumble

- imqwerty

:O

- imqwerty

༼•͟ ͜ •༽

- mathmate

Hi, This is what I have.

- imqwerty

hi, i like it what u have :)

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.