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Would anyone have the answers

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Would anyone have the answers [#permalink] New post 22 Oct 2009, 13:47
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Hi, I came across the following questions in a series of GMAT permutation questions, and they stumped me (either I couldn't figure them out, or I thought the proposed answer was wrong):

1. In how many ways can 5 people sit in a circle:
1) without any restrictions: got this (5-1)!
2) one should not have the same neighbors in any two arrangements: not sure about this, but have 4! - 3*5 = 9

2. In how many ways can 5 identical black balls and 7 identical white balls be arranged in a row so that no two black balls are together:
A) 56
B) 64
C) 65
D) 316
E) 560
The answer key says A, but I'd like to understand the mechanism

3. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits presents:
A) 13^4 * 48 * 47
B) 13^4 * 24 * 47
C) 48 Choose 6
D) 13^4
E) 13^4 * (48 Choose 6)
The answer key says B, but I worked it out and A makes more sense

4. How many numbers greater than a thousand can be made using the following digits without repetition: 1, 0, 3, 4, and 5:
A) 96^2
B) 96*2
C) 576
D) 24^2
E) 24*96
This one makes no sense to me, the answer may have been written incorrectly, because any combination of these digits will be greater than a thousand, even if we start with 0, so my answer would have been 5!=120

5. A mocktail has to be prepared from three kinds of whiskey, 4 kinds of soda and five kinds of fruit juice. How many mocktails can be made taking at least one of each kind?
A) 235
B) 325
C) 3255
D) 5235
E) 565
Again, another example of a poorly written question. From the question stem, what other possibilities are there other than multiply the choices through, i.e. 3*4*5 = 60, which is not an answer option

Any guidance would be good. Thanks!
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 11:07
These are some serious questions dude.

I am only confident abt the 4th answer

Solution
You started of correctly -
total number of 5 digit numbers = !5-!4(when 0 is in the ten thousandth place)
Total number of 4 digit numbers excluding 0 = !4
Total number of 4 digit numbers excluding 1 = !4-!3(when 0 is in the thousands place)
similarly for 3,4 & 5

Adding all these 120-24+24+4*(24-6) = 192 = 96*2 hence answer is B.
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 11:11
Seshouan wrote:
5. A mocktail has to be prepared from three kinds of whiskey, 4 kinds of soda and five kinds of fruit juice. How many mocktails can be made taking at least one of each kind?
A) 235
B) 325
C) 3255
D) 5235
E) 565
Again, another example of a poorly written question. From the question stem, what other possibilities are there other than multiply the choices through, i.e. 3*4*5 = 60, which is not an answer option

Any guidance would be good. Thanks!

Aren't Mocktails non alcoholic :?:

Anyways, in your solution you considered one of each kind instead of at least 1 of each kind. I am still trying to figure out the answer.
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 12:01
Seshouan wrote:
2. In how many ways can 5 identical black balls and 7 identical white balls be arranged in a row so that no two black balls are together:
A) 56
B) 64
C) 65
D) 316
E) 560
The answer key says A, but I'd like to understand the mechanism


This is nearly impossible to explain but this might point you in the right direction -
take 5 black balls
put 4 white balls between them
now you have 3 white balls remaining
start by putting all 3 in the left most position
now start moving all 3 towards right
you can put all 3 together in 6 places
now get all 3 back to the left split it into 2 leftmost and 1 free
start moving the free one towards right
you can do it in 5 ways
get the 1 free back left
now split it into 1 left 2 free
again 5 ways
when 2 are together
split that 2 again and move right and so on
you will see a pattern

6+5+5+4+3+2+1
+4+4+3+2+1
+3+3+2+1
+2+2+1
+1+1
= 6+2*5+3*4+4*3+5*2+6*1 = 56

This is as bad as counting each scenario but I couldn't find a better solution.
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 13:00
atish wrote:
These are some serious questions dude.

I am only confident abt the 4th answer

Solution
You started of correctly -
total number of 5 digit numbers = !5-!4(when 0 is in the ten thousandth place)
Total number of 4 digit numbers excluding 0 = !4
Total number of 4 digit numbers excluding 1 = !4-!3(when 0 is in the thousands place)
similarly for 3,4 & 5

Adding all these 120-24+24+4*(24-6) = 192 = 96*2 hence answer is B.

Atish why don't you try normal permutation method :wink:
4 digit nos = 4*4*3*2 = 96
5 digit nos = 4*4*3*2*1 = 96
total possible nos = 96*2...OA B
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 13:12
Q5. A mocktail has to be prepared from three kinds of whiskey, 4 kinds of soda and five kinds of fruit juice. How many mocktails can be made taking at least one of each kind?
A) 235
B) 325
C) 3255
D) 5235
E) 565

u need to know following ri=ule in combination:
if you have n different things and u r allowded to choose any no of things then u can choose in 2^n ways ( including a choice of not selecting)
2^n = nC0+nC1+nC2+...........+nCn
hence no of ways for selecting at leaset one will be 2^n - 1

use this for our mocktails ( or cocktails :lol: :wink: ) with 3W, 4S, 5J
this can be done in (2^3-1)*(2^4-1)*(2^5-1) = 7*15*31 = 3255 ways OA C

i hope this helps :wink:
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Re: Would anyone have the answers [#permalink] New post 23 Oct 2009, 13:14
bhushan252 wrote:
Atish why don't you try normal permutation method :wink:
4 digit nos = 4*4*3*2 = 96
5 digit nos = 4*4*3*2*1 = 96
total possible nos = 96*2...OA B


Define normal permutation method :-D , I don't see permutations in your answer, just a bunch of factorials and multiplications with minimal explanation.
If there is a predefined formula for such questions I am all ears. Does it take 0 and non 0 numbers into account?

Appreciate your advice.
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Re: Would anyone have the answers [#permalink] New post 09 May 2011, 00:00
Interesting ones.
Last one uses good concept.
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Re: Would anyone have the answers [#permalink] New post 29 Oct 2011, 05:21
Could you plz explain question 2?
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Re: Would anyone have the answers [#permalink] New post 01 Nov 2011, 04:26
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1.1) i like to write is as \frac{5!}{5} = 4! = 24
1.2) AxB or BxA, so same neighbours are counted twice. that's why ans = \frac{4!}{2} = 12

2) A
arrange the 7 white balls first. _W_W_W_W_W_W_W_
5 black balls have 8 spaces inbetween, so arrangements = \frac{8*7*6*5*4}{5!} = 56

3) B
\frac{13^4*48*47}{2} .. we divide by two to avoid repetition of the last two cards i.e. ab or ba are the same in this context.

4) B
we have to divide it into two cases i.e. 4-digit and 5-digit because 1034 and 10345 are both greater than 1000 and we can't put a condition that 0 can't be placed on 2nd position.
4-digit: 4*4*3*2 = 96 (0 can't be the first digit)
5-digit: 4*4*3*2*1 = 96 (0 can't be the first digit)
total = 96*2

5) C
at least one out of n is 2^n-1.
total mocktails = at least one whiskey * at least one soda * at least one juice
= 2^3-1*2^4-1*2^5-1

=3255
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Re: Would anyone have the answers [#permalink] New post 01 Nov 2011, 05:57
MBAhereIcome wrote:
1.1) i like to write is as \frac{5!}{5} = 4! = 24
1.2) AxB or BxA, so same neighbours are counted twice. that's why ans = \frac{4!}{2} = 12

2) A
arrange the 7 white balls first. _W_W_W_W_W_W_W_
5 black balls have 8 spaces inbetween, so arrangements = \frac{8*7*6*5*4}{5!} = 56

3) B
\frac{13^4*48*47}{2} .. we divide by two to avoid repetition of the last two cards i.e. ab or ba are the same in this context.

4) B
we have to divide it into two cases i.e. 4-digit and 5-digit because 1034 and 10345 are both greater than 1000 and we can't put a condition that 0 can't be placed on 2nd position.
4-digit: 4*4*3*2 = 96 (0 can't be the first digit)
5-digit: 4*4*3*2*1 = 96 (0 can't be the first digit)
total = 96*2

5) C
at least one out of n is 2^n-1.
total mocktails = at least one whiskey * at least one soda * at least one juice
= 2^3-1*2^4-1*2^5-1

=3255



Thanks a lot MBAhereIcome
BTW, how was your test? I see you mentioned that your GMAT test was supposed to be on 10-31-2011
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Re: Would anyone have the answers [#permalink] New post 01 Nov 2011, 06:16
shahideh wrote:
Thanks a lot MBAhereIcome
BTW, how was your test? I see you mentioned that your GMAT test was supposed to be on 10-31-2011


you're welcome. i need to change my gmat date; i haven't taken it yet :(
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Re: Would anyone have the answers   [#permalink] 01 Nov 2011, 06:16
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