CAT 2008: Quantitative Questions a Day 134 -Till end -> The Discussions

smileyface

Quote:

Originally Posted by smileyface

QQAD 147: CAT 2008, Arrangements

Is 6*5*4*3 = 360 , the answer ? It seems very simple to me , don't know if I'm missing out on something

Oh , ok.. I realised my mistake but I didn't understand the above explaination. Please elaborate.

awadhesh016

no its not dat simple.have a look @ ur 11 th class permutation & combination book.u'll find what u r missing.

vivekr

I've a slightly different approach to solving this problem.
The question can be considered to be the number of ways of selecting the numbers (1,2,3,4,5,6) each appearing from 0 to 4 times.
So, n=4, r=6 and we can use our (n+r-1)C(r-1) funda to get 10C5 = 126.
=> Option (c)

awadhesh016

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but it is a stub and does not seem to give a true picture of pagalguy.com
please help improve it.

serena

if the order is unimportant then the numbers on the dice can be arranged in 4! ways and the dice themselves can be thrown in 4 ways.
therefore i think the number of different arrangements are 4*4! ways or 96 ways.
therefore the answer is option (a) . i think.

serena

horrible at permutations and combinations could someone please help and explain how exactly to tackle these kind of problems.

slam

Quote:

Originally Posted by implex

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Quantitative Question # 147
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Four dices(6-faced) are thrown. How many different arrangements can one get if the order is unimportant(e.g. 2356 is same as 2635 or 1124 is same as 4211)?

(a) 96 (b) 108 (c) 126 (d) 156 (e) none of these

4 same => 6c1 = 6 ways
3 same => 6c1 * 5c1 = 30 ways
2 same, 2 same => 6c2 = 15 ways
2 same, 2 diff => 6c1 * 5c2 = 60 ways
4 diff => 6c4 = 15 ways. => total = 126 ways.

slam

Quote:

Originally Posted by vivekr

I've a slightly different approach to solving this problem.
The question can be considered to be the number of ways of selecting the numbers (1,2,3,4,5,6) each appearing from 0 to 4 times.
So, n=4, r=6 and we can use our (n+r-1)C(r-1) funda to get 10C5 = 126.
=> Option (c)

I think you meant 9c5 = 126. Anyway, that's a different way to do the question!

singh_kanwar

Quote:

Originally Posted by vivekr

I've a slightly different approach to solving this problem.
The question can be considered to be the number of ways of selecting the numbers (1,2,3,4,5,6) each appearing from 0 to 4 times.
So, n=4, r=6 and we can use our (n+r-1)C(r-1) funda to get 10C5 = 126.
=> Option (c)

sir how is n=4??? if u r saying dat "each appearing frm 0 to 4 times".....dat sud mean n=5??

singh_kanwar

after beating my head for arnd 1 hr i m wid this
if the order is unimportant n 1124 is same as 4211 den
the possibility of nos 1,2,3,4,5,6 can appear on the 4 dice in the following ways:

all dice having the same nos
=> 6c1 = 6 ways
having 3 same nos
=> 6c1 * 5c1 = 30 ways
having 2 same each
=> 6c2 = 15 ways
having 2 same& 2 diff nos
=> 6c1 * 5c2 = 60 ways
all 4 diff nos
=> 6c4 = 15 ways.
=> total = 126 ways.

so the answer is (c) 126