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

vaibhaviitd

Lets look at it case-wise

Case I: Same number on all 4 dice - No. of ways = 6C1 = 6

Case II: Exactly 2 numbers out of 6 on all 4 dice :
Possible Arrangements if nos are a,b - aaab,aabb,abbb
No. of ways - 6C2 * 3 = 45

Case III: Exactly 3 numbers out of 6 on all 4 dice:
Possible Arrangements if nos are a,b,c - aabc,abbc,abcc
No. of ways - 6C3 * 3 = 60

Case IV: Different numbers on all the dice: No. of ways = 6C4 = 15

Therefore, total no. of ways = 6+45+60+15 = 126

Hence option (c)

rigved123

The answer to QQAD 147 is 126.

prvineeth · 02:51 PM

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Quantitative Question # 147
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Ans:

When all 4 are same:

no of cases = 6

when 3 are same

no of cases=6 * 5

when 2 are same

no of cases=6C2+6 * 5C2=75

when all different

no of cases=6 C4=15

So total cases = 6+30+75+15 =126

answer option (c) 126

priyankanikki

the answer will be 96

jayaram85

hi implex

as usual i have some doubts i agree with the first case of 6c1 ways

but lets take a simple case of 1,2 being the only numbers the various combinations possible for 3 1s and 1 2 and 3 2s and 1 1 can be

1112
1121
1211
2111
2221
2212
2122
1222 so in all 8 ways for 2 numbers taken one of them as once and the other thrice

similarly for 1,3 1,4 1,5 etcc... upto 5,6 we get 15*8 so total 120 ways

implex any wrong thought process????

shivam_01

Quote:

Question 147
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

Moi take ::

case 1: When all 4 are same digits then :
no of combinations = 6

case 2: When 3 are same digits then :
no of combinations = 6* 5 = 30

case 3: When 2 are same digits then :
no of combinations = 6C2+6 *5C2 = 75

case 4: When all are different digits then :
no of combinations =6 C4=15

So total = 6+30+75+15 =126

or we can use
n+r-1 C r-1 = 9C5 = 126 ways

answer option (c) 126

implex

Quote:

Originally Posted by jayaram85

hi implex

as usual i have some doubts i agree with the first case of 6c1 ways

but lets take a simple case of 1,2 being the only numbers the various combinations possible for 3 1s and 1 2 and 3 2s and 1 1 can be

1112
1121
1211
2111
2221
2212
2122
1222 so in all 8 ways for 2 numbers taken one of them as once and the other thrice

similarly for 1,3 1,4 1,5 etcc... upto 5,6 we get 15*8 so total 120 ways

implex any wrong thought process????

The 8 you have shown will count as 2

the question clearly says we have to count combinations and not permutations
order does not matter

Read the problem again !!! Its correct and unambiguous

jayaram85

hi puys

let me thank implex first for pointing out my mistake
now ill try my answer

case 1 all are same so 6 cases 1111, 2222, 3333, 4444, 5555, 6666 6 ways

case 2 3 are same one different here arrangement doesnt matter thats wat implex pointed out example 1112 is same as 1121, 1211, 2111 so only one case similarly 2221, 2212, 2122, 1222 are same so for every combination 0f 2 different numbers there are 2 ways number of ways of selecting 2 numbers is 6c2=15 thus total is 15*2=30

case 3 2 equal 2 equal number of ways of selecting 2 numbers is 6c2=15
case 4 2 equal 2 different number of ways of selecting the equal number is 6c1 from the remaining we have to select any 2 which is 5c2

so total ways are 6c1* 5c2= 60

case 5 all are different simplest case 6c4= 15

so total is 126

option c)

Varun Khullar

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Quantitative Question # 148
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100 numbers are written around a circle. The sum of every 8 consecutive numbers is 25/2. The 9th number is -1/2, the 19th number is 3/4 and 20th number is 2. What is the 50th number?

(a) 4 (b) 5 (c) 1 (d) 3 (e) none of these

The question was that was 148th Last Year
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A car travels downhill at 72 kmph (kilometers per hour), on the level at 63 kmph, and uphill at only 56 kmph The car takes 4 hours to travel from town A to town B. The return trip takes 40 minutes more. What is the distance between the two towns in kilometers?

(1) can not be determined (2) 191 (3) 255 (4) 273 (5) none of these

tyro.novice

QQAD 148 AS the numbers are arranged around a circle, numbers will be repeated and the sum of every 8 consecutive numbers is same
so 1, 9,17,25 etc . are same
if we consider the series from 17 th to 24th number
17th , 21 same as 9th
take 18, 22 as x
19,23 same
20,24 same
sum=25/2
hence
4+3/2+2x-1=25/2
x=4
50 th term is equal to the 18 th term
option a
4