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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

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. _________________

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. _________________

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

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 4 digit nos = 4*4*3*2 = 96 5 digit nos = 4*4*3*2*1 = 96 total possible nos = 96*2...OA B _________________

Bhushan S. If you like my post....Consider it for Kudos

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 ) 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 _________________

Bhushan S. If you like my post....Consider it for Kudos

Atish why don't you try normal permutation method 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 , 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?

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 \(= 2^3-1*2^4-1*2^5-1\)

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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