CAT 2007: Quantitative Questions a Day 101 to 133 - The Discussions

[vineet.nitd]

Please carry the QQAD discussion for the next 33 problems here.

This would be the penultimate thread for QQAD this year

Good Luck to all

[Aarav]

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Quantitative Question # 101 (the answer for both Part A and Part B are in the same set of options)
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Question(s):

(Part A)
An insect starts at vertex A of a certain cube and is trying to reach at vertex B, which is opposite A, in 5 or fewer steps where a step consists travelling along an edge from one vertex to another. The insect will stop as soon as it reaches B. The number of ways in which the insect can achieve its objective is

(Part B)
A committee of 5 is to be chosen from a group of 9 people. How many ways it can be done if Vikram and kaizen must serve together or none at all, and Sunit and Pratyush refuse to serve with each other?

(a) 40 (b) 36 (c) 41 (d) 30 (e) 48

[sunit_chotu]

Part A : See the figure attached: The insect can go from A to B either in exactly 3 steps or in exactly 5 steps.
No: of ways to go from A to B in exactly 3 steps = (3C1)*(2C1) = 6 ways.

because at vertex A it has to choose 1 edge out of 3 which it will do in 3C1 ways. and at each vertex 1 or 3 or 5 it has to choose 1 edge out of 2 edges which it will do in 2C1 ways.

No: of ways to go from A to B in exactly 5 steps = No: of ways to go from A to B in exactly 5 steps when no edge is traversed more than once + No: of ways to go from A to B in exactly 5 steps when one of the edge is traversed more than once.

No: of ways to go from A to B in exactly 5 steps when no edge is traversed more than once = 6 ways.

now lets find out the number of ways when insect goes from A - 1 - A, that is it has completed 2 steps and is back to A from which it has to reach B in remaining 3 steps which it can do in 6 ways.

so when edge A-1 is repeated = 6 ways. when edge A3 is repeated = 6 ways, when edge A-5 is repeated = 6 ways.
hence (6+6+6) = 18 ways
now say the insect go from vertex A to any of the three vertex . lets say it goes from A to 1 here when edge 1-2 is repeated then insect has 2 ways and when 1-6 is repeated insect has 2 ways. so total 4 ways, hence 3C1*4 = 12 ways.

one more case is there when insect travels from A-1-2-3-2-B i.e third side is repeated here also 6 ways will be there.

hence total number of ways = 6 + 6 + 18 + 12 + 6 = 48 ways
option (e) 48 ways.

[sunit_chotu]

Part B :

Case 1 : When Vikram and Kaizen both are chosen.

now we are left with 7 people out of which we have to chose 3 people.

Case A1 : when Pratyush is chosen Sunit cannot be chosen
so remaining 2 can be chosen from remaining 5 in 5C2 ways.

Case A2 : when Sunit is chosen Pratyush cannot be chosen
so remaining 2 can be chosen from remaining 5 in 5C2 ways.

Case A3 : when both Sunit and Prastyush are not chosen
so remaining remaining 3 can be chosen in 5C3 ways

So total number of ways in this case = 5C2 + 5C2 + 5C3 = 10 + 10 + 10 = 30 ways.

Case B : When Vikram and Kaizen both are not chosen .
We are left with 7 people out of which we have to chose 5 people.
Case B1 : when Pratyush is chosen Sunit cannot be chosen
so remaining 4 can be chosen from remaining 5 in 5C4 ways.
Case B2 : when Sunit is chosen Pratyush cannot be chosen
so remaining 4 can be chosen from remaining 5 in 5C4 ways.
So total number of ways in this case = 5C4+5C4 = 10

Case C : when all the 4 are not chosen then 5 people can be chosen out of 5 in 5C5 ways = 1 way

Hence total number of ways = 30 + 10 + 1 = 41 ways.
option ( c) 41 ways.

[mehtamanas]

PART B ...Ans : 5C2*2C1 + 2C1*5C4 + 5C3 + 5C5 = 41 Ways...

[marijuana_user]

Originally Posted by sunit_chotu

Part B :

Case 1 : When Vikram and Kaizen both are chosen.

now we are left with 7 people out of which we have to chose 3 people.

Case A1 : when Pratyush is chosen Sunit cannot be chosen
so remaining 2 can be chosen from remaining 5 in 5C2 ways.

Case A2 : when Sunit is chosen Pratyush cannot be chosen
so remaining 2 can be chosen from remaining 5 in 5C2 ways.

Case A3 : when both Sunit and Prastyush are not chosen
so remaining remaining 3 can be chosen in 5C3 ways

So total number of ways in this case = 5C2 + 5C2 + 5C3 = 10 + 10 + 10 = 30 ways.

Case B : When Vikram and Kaizen both are not chosen .
We are left with 7 people out of which we have to chose 5 people.
Case B1 : when Pratyush is chosen Sunit cannot be chosen
so remaining 4 can be chosen from remaining 5 in 5C4 ways.
Case B2 : when Sunit is chosen Pratyush cannot be chosen
so remaining 4 can be chosen from remaining 5 in 5C4 ways.
So total number of ways in this case = 5C4+5C4 = 10


Hence total number of ways = 30 + 10 = 40 ways.
option ( a) 40 ways.

hey!

wat about the part that none among the four is chosen::
then we have to select 5 out of 5 people...that is 1 more way....

hence total ways::41....

regards!

[marijuana_user]

Originally Posted by sunit_chotu

Part A : See the figure attached:

cant c any...i think u forgot to attach it...

IS it that,if v r saying that a corner is opposite to another corner,then it means that B is diagonally opposite to A.....????

[sunit_chotu]

Originally Posted by marijuana_user

hey!

wat about the part that none among the four is chosen::
then we have to select 5 out of 5 people...that is 1 more way....

hence total ways::41....

regards!

THANKS was in a hurry had some work in office so missed out this case . Have edited my soln.

[sunit_chotu]

Originally Posted by marijuana_user

cant c any...i think u forgot to attach it...

IS it that,if v r saying that a corner is opposite to another corner,then it means that B is diagonally opposite to A.....????

Ya had forgot to attach it, see that post again i have already attached it. yes A and B are diagonally opposite, the diagonal is the longest diagonal of the cube not of a side of the cube.

[krsh.vik]

Part B
none of them 5c5
vk 5c3
vks 5c2
vkp 5c2
s 5c4
p 5c4

vkps stand for vikram,kaizen,pratyush and sunit respectively
total 41