Permo Combo continues 5

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Sun Aug 09, 2009 10:09 am

Each of the three dice has its six faces numbered respectively 1,2,3,4,5,6 but the dice themselves are of different colours. if three dice are simultaneously thrown out of the dice box in how many of these way they fall in such a way that two of the dice show same number and the third a different number.

90
120
210
216
1296

Charged up again to beat the beast

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Source: Beat The GMAT — Problem Solving | Sun Aug 09, 2009 10:09 am

real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Sun Aug 09, 2009 12:00 pm

is it 90?

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DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

by DanaJ » Sun Aug 09, 2009 12:03 pm

Since I did guess the previous permutations question I attempted, I'll try this one too.

First off let's not think of the dice as colored. Consider three identical dice and you're trying to find out the ways in which two hit the same number while the third hits a different one. There are 6 numbers that the two dice can show together: 1 through 6. For each of the 6 possibilities, the third dice has to show 5 other numbers (EXCEPT the one shown by the other two). In total, you have 6*5 = 30 possibilities.

But the dice are also colored, which adds an extra twist: in this case, there will be 3C2 = 3 possible combinations of the two "common" dice. The third dice doesn't matter now, since it will be the one left.

This makes my guess 30*3 = 90.

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ankitns: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Jul 13, 2009 8:58 pm; Thanked: 10 times; GMAT Score:710

by ankitns » Sun Aug 09, 2009 12:12 pm

I agree. Derived 90 via same approach.

What is the OA?

Cheers.

DanaJ wrote:Since I did guess the previous permutations question I attempted, I'll try this one too.

First off let's not think of the dice as colored. Consider three identical dice and you're trying to find out the ways in which two hit the same number while the third hits a different one. There are 6 numbers that the two dice can show together: 1 through 6. For each of the 6 possibilities, the third dice has to show 5 other numbers (EXCEPT the one shown by the other two). In total, you have 6*5 = 30 possibilities.

But the dice are also colored, which adds an extra twist: in this case, there will be 3C2 = 3 possible combinations of the two "common" dice. The third dice doesn't matter now, since it will be the one left.

This makes my guess 30*3 = 90.

Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Sun Aug 09, 2009 1:29 pm

OA is 90 but I m yet not able to comprehend the logic

Charged up again to beat the beast

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ankitns: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Mon Jul 13, 2009 8:58 pm; Thanked: 10 times; GMAT Score:710

by ankitns » Sun Aug 09, 2009 1:36 pm

Read this thread..it may help...focus on the part B of the solution...

https://www.beatthegmat.com/dice-t42198. ... light=dice

Cheers.

maihuna wrote:OA is 90 but I m yet not able to comprehend the logic

Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!

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prindaroy: Master | Next Rank: 500 Posts; Posts: 113; Joined: Thu Jul 16, 2009 11:23 am; Thanked: 15 times; GMAT Score:730

by prindaroy » Sun Aug 09, 2009 6:55 pm

Blue.............Red....................Pink
6 poss..........1 poss................5 poss. = 6 * 5 * 1 = 30

5 poss..........6 poss................1 poss. = 5 * 6 * 1 = 30

1 poss..........5 poss................6 poss. = 1 * 5 * 6 = 30

30 + 30 + 30 = 90

Think of it this way; each color will either have 5 possibilities or 1 possibility or 6 possibilities., so you should account for all three scenarios

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winnerhere: Master | Next Rank: 500 Posts; Posts: 231; Joined: Thu Apr 12, 2007 2:45 amFollowed by: 1 member; Thanked: 5 times

by winnerhere » Wed Aug 12, 2009 5:57 am

maihuna,

I got 90 by this logic(no need to think about the colour at all)

there are 5 different possibilities of getting same number twice

(1,1,x) (2,2,x)........(6,6,x)

Number of different permutations of (1,1,x) is 3!/2!

formula used : number of different arrangements of n items with r similar items is (n!/r!)

x can take 5 different values - so (3!/2!) * 5 = 15

for (1,1,x) it is 15..

for all the six possibilities it is 6 * 15 = 90

dont confuse by colours......because anyway (1,2) is different from (2,1) - even if u have the same colour dice.

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