Few Questions

[sporty.arjun]

Q.how many 5 digit nos. can be formed using the digits 0 2 3 4 5 when repetition is allowed such that the number is divisible by 2 or 5 or both?
OA:1500
Q. 4 persons enter a lift of a 7 storey building at the grnd flr. in how many ways can they get out of the lift other than the ground floor?
OA:1296
Q.10 diff alphabets r given. 2 of these letters followed by 2 digits are used to number the products of a company. in how many ways can the products be numbered?
OA:1000
Q.P(2n+1,n-1):P(2n-1,n)=3:5, find n...{what is this representation}?
cud sum1 tell me how 2 solve these?

[mystic87]

#1.
first digit has 4 choices (2,3,4,5)
second, third, fourth digits each has 5 choices (all 5)
fifth digit can only have even or 0 or 5 ( all but 3) so that makes 4 choices

4x5x5x5x4 =2000

#2. working on it...

#3. 10x10x10x10=10000

#4. P(k,r) = kPr <--- Permutation where k=total number, r=selected number

[Goldust]

I would think that for the second question, each person could choose to get out on any of the 6 floors (after excluding the ground floor). Hence, the total possible ways = 6*6*6*6 = 1296.

[sporty.arjun]

hey mystic.....in d 1st sum 4*5*5*5*4 is 2000 not 1500..and in d 3rd sum 10*10*10*10 is 10000 not 1000...

[lookingahead]

yes true exactly my point u had me stumped here 4*5*5*5*4 = 2000

[chestnut.cc]

Q.how many 5 digit nos. can be formed using the digits 0 2 3 4 5 when repetition is allowed such that the number is divisible by 2 or 5 or both?

If the question means 2 and 5 or both... 4*5*5*5*2 + 4*5*5*5*1... 1500

[grexam]

1)can u please explain third problem?

2)Are questions like that mentioned in fourth question above also given?

[prithibi]

4. answer: n=4
(2n+1)P(n-1): (2n-1)P(n)=3:5
=>(2n+1)!/(2n+1-n+1)!: (2n-1)!(2n-1-n)!=3/5
=>(2n+1)!/(n+2)!: (2n-1)!/(n-1)!=3/5
=>(2n+1)!(n-1)!/(n+2)!(2n-1)!=3/5
=>(2n+1)2n/(n+2)(n+1)n=3/5
=>3n^2-11n-4=0
n= 4 or -1/3
n can not take negative value and n= 4
got it?

[prithibi]

3. explanation
10 letters out of which 2 are selected and 10 digits out of which 2 digits are selected to make the code i. e. the code consists of four places. each place is occupied by 10 possible outcomes since repetition is allowed. thats why the four places make 10*10*10*10= 10000 possible outcomes

[mystic87]

Originally Posted by chestnut.cc

Q.how many 5 digit nos. can be formed using the digits 0 2 3 4 5 when repetition is allowed such that the number is divisible by 2 or 5 or both?

If the question means 2 and 5 or both... 4*5*5*5*2 + 4*5*5*5*1... 1500

Sorry for the miscalculation for this q. But can someone explain why #1 shouldn't be 4*5*5*5*4 if it's asking for "divisible by 2 or 5 or both?"

Can someone remind me the shortcut to simplify 3N^2 - 11n - 4 to get n= 4 or -1/3? I know this is highschool math...but it's been a few years and i really can't remember.