rosala

thank you so much Melody for taking the time to answer my questions,

i'd like to start wiith the first one.

the answer that youve given to question no. 1 , thats not the answer thats present in my book. the answer to it is (x-1 )^x  - 1 , my doubt in this was that why are we subtracting one here? the (x-1)^x ca efro forula (n)^r.

2.oh Melody i feel bad to have troubled you for this problem , i was just making 2 silly mistakes. but couldnt see them,thank you for  the answer and explanation. 

3. 

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less than      200,000,000   only containing the digits 1 and 2
1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1

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your third answer is correct btw!

i have no idea what you did here...otherwise i understand everything after that..... could you pls explain this part ? thank you for the answers! 

CPhill i think theres some problem in the 'c' parts answer becoz my teachers answer is 144.

its somewhat like:

No. of ways in which none of U , A , E are together = 3! x ⁴P₃  = 3! x 4!/1! = 3! x 4! = 144.

i have no idea what is done here.

dont you think that U, A , E ..that are vowels...and when they wont appear together, then the answer will be the same as the 'b' part???? 

thanks hectictar for noticing..i appreciate it!thank you! 

CPhill i have no idea about it!? 

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Can  put these in any one of 5 positions    12, 23, 34, 45, 56

And for each of these, there are 4!  ways to arrange the other letters

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what do you mean byt he first line?i thought there were 5! ways to arrange them???

can you pls make it a little simple....i really cant see the permutations in them....its too confusing!Thanks!

sorry i get the varying part ...just other is confusing....its me rosala...i was just too lazy to log in....lol

Nice answer CPhill...i just dont get the par twhere you used "a"....what does that mena?could you write it in a longer form....

my teacher's answer is in this way:

1!+2!+3!+4!+5!/3

=3+3!+4!+5!/3

+3(1+ 1x2 + 1x2x4 + 1x2x4x5)/3

3!, 4! .......n! all have a factory of 3

1!+2!=3

therefore, ....... is divisible by 3.

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i asked him about the answer and he said that those added terms will extend to n , tho i dont understand how to write them by taking 3 as a factor,.....could you pls solve it like that.!!!! Thank you so much CPhill....

Heres an energy frink for you....to help you answer my questions!thanks.

CPhill...i have no idea about your answer......shouldnt n vary from 1 to 43?

*Note that in every case, n  will divide both terms evenly  because n will be a factor of both terms*

what do you mean by that statement..

i dont understand anything after that either....pls could you elaborate.....thank you!

Wow Steve , that thing is so lit, good job.....i was so amazed to see it.... i think admin would love it....awesome work!

Hi Hectictar, happy birthday.....i was actually at a bakery to buy your cake but then CPhill told me he already bought you one...so i thought to buy you gifts from all of us...your beloved forum....

here they are....hope you like them!

Hope you have a blasting day......happy birthday !

something out of my reach apparently