Challenge Problems

Expert replies

darknight: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Nov 04, 2009 5:17 am

by darknight » Wed Nov 04, 2009 5:59 am

Hi,

Please find below questions to the problems that I could not solve. The answers are given below but I'm not aware how the answer is arrived at. So please if you can post a solution to these problems it would be a great help.

Cheers,
darknight

[[THIS IS THE TEXT VERSION. I WOULD REQUEST YOU TO PLEASE HAVE A LOOK AT THE IMAGE BELOW THE TEXT VERION FOR BETTER COMPREHENSION.]]

1. If 2nC3: nC2 = 44:3 find n. (Ans: 6)
2. If nC12 = nC8 what is the value of 22Cn. (Ans: 231)
3. If 4 times the number of permutations of n things 3 together is equal to 5 times the number of permutations of n-1 things 3 together find n. (Ans: 15)
4. A boat is to be manned by 8 men of whom only 2 can row on bow side and only 1 can row on stroke side. In how many ways can the crew be arranged? (Ans: 5760)
5. An 8-oared boat is to be manned by a crew chosen from 11 men out of which 3 can steer but cant row and the rest can row but not steer. In how many ways can the crew be arranged if 2 of the men who can row can only do so on the bow side? (Ans: 25920)
6. In how many ways can 5 prizes be given to 4 people when each is eligible for one or more prizes? (Ans: 1024)

Source: Beat The GMAT — Quantitative Reasoning | Wed Nov 04, 2009 5:59 am

fireplayer: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Fri May 08, 2009 8:34 pm

by fireplayer » Sun Nov 08, 2009 8:22 am

i am confused with the last question, how can be the ans= 1024 when everyone eligible to get one or more than one prize? someone will get none if other get 3 prize?

vyomb: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Thu Feb 26, 2009 1:19 am; Thanked: 3 times; GMAT Score:650

by vyomb » Sun Nov 08, 2009 9:20 am

a single person can have the following options
amount of prize received=0 or 1 or 2 or 3 or 4
so total 5 possibilities.

hence for 4 people we have 4 pow 5=1024 ways.

Vinod R: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Nov 09, 2009 2:57 am; Thanked: 1 times

by Vinod R » Mon Nov 09, 2009 4:16 am

Hi,
Solutions for 3 questions among the given set of questuions are as follows:
1. If 2nC3: nC2 = 44:3 find n.

Sol: Expand the two 2nC3 and nC2 with standard formulae:
2n!/3!*(2n-3)! /n!/2!*(n-2)! = 44/3
2n(2n-1)(2n-2)(2n-3)!/3!*(2n-3)!/n*(n-1)*(n-2)!/2!*(n-2)! = 44/3
Now (2n-3)! and (n-2)! will be removed.

2n(2n-1)(2n-2)*2!/n*(n-1)*3! = 44/3
Solving the above u will get
2n-1 = 11
n =6.

2. If nC12 = nC8 what is the value of 22Cn. (Ans: 231)

Sol: Solution to this problem is quite simple as it is based on the standard formulae :
nCr = nCn-r

Just make a simple guess :
For satisfying the above standard formulae "20" will be the right choice. So substitute 20 then

20C12 = 20C8
So 22C20 = 231

3. If 4 times the number of permutations of n things 3 together is equal to 5 times the number of permutations of n-1 things 3 together find n.

Sol: The above question can be written as:

4(nP3)= 5.((n-1)P3)

By solving the above u wil get:
4n = 5n - 15;
n = 15.

Regards,
Vinod.

darknight: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Nov 04, 2009 5:17 am

need clarification

by darknight » Wed Nov 11, 2009 9:23 pm

vyomb wrote:a single person can have the following options
amount of prize received=0 or 1 or 2 or 3 or 4
so total 5 possibilities.

hence for 4 people we have 4 pow 5=1024 ways.

Thanks vyomb.

But it is said that each is eligible for 1 or more prize. will this not act as a constraint for taking no person getting a prize?

Thanks again,
darknight

darknight: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Nov 04, 2009 5:17 am

by darknight » Wed Nov 11, 2009 9:25 pm

Vinod R wrote:Hi,
Solutions for 3 questions among the given set of questuions are as follows:
1. If 2nC3: nC2 = 44:3 find n.

Sol: Expand the two 2nC3 and nC2 with standard formulae:
2n!/3!*(2n-3)! /n!/2!*(n-2)! = 44/3
2n(2n-1)(2n-2)(2n-3)!/3!*(2n-3)!/n*(n-1)*(n-2)!/2!*(n-2)! = 44/3
Now (2n-3)! and (n-2)! will be removed.

2n(2n-1)(2n-2)*2!/n*(n-1)*3! = 44/3
Solving the above u will get
2n-1 = 11
n =6.

2. If nC12 = nC8 what is the value of 22Cn. (Ans: 231)

Sol: Solution to this problem is quite simple as it is based on the standard formulae :
nCr = nCn-r

Just make a simple guess :
For satisfying the above standard formulae "20" will be the right choice. So substitute 20 then

20C12 = 20C8
So 22C20 = 231

3. If 4 times the number of permutations of n things 3 together is equal to 5 times the number of permutations of n-1 things 3 together find n.

Sol: The above question can be written as:

4(nP3)= 5.((n-1)P3)

By solving the above u wil get:
4n = 5n - 15;
n = 15.

Regards,
Vinod.

Thanks Vinod for your reply. It certainly made my day. If possible can u pls try a hand @ the boat problems?

Thanks again,
darknight

darknight: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed Nov 04, 2009 5:17 am

by darknight » Wed Nov 11, 2009 9:51 pm

fireplayer wrote:i am confused with the last question, how can be the ans= 1024 when everyone eligible to get one or more than one prize? someone will get none if other get 3 prize?

thanks for your reply, fireplayer.

check out vyomb's reply to ur question in this thread.

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