|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 18 Oct 2004
Posts: 9
Followers: 0
Kudos [?]:
0
[0], given: 0
|
John has 7 bananas and 3 kiwis. In how many ways can John [#permalink]
 01 Feb 2005, 19:23
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
John has 7 bananas and 3 kiwis. In how many ways can John divide the 10 fruit between two parcels, if there has to be an equal total number of fruit in either parcel and so that there is at least one kiwi in each parcel.
Pls. show your work.
Thanks.
|
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
We need to select 5 from 10, with at least 1 kiwi.
1 kiwi
C(3,1)C(7,4)=3*7*6*5/6=105
And that's it. No need to do two kiwis since it is symmetric.
|
|
|
|
|
|
SVP
Joined: 30 Sep 2004
Posts: 1548
Location: Germany
Followers: 4
Kudos [?]:
15
[0], given: 0
|
1. 1 k + 4 b or 2 k + 3 b
3C1 * 7C4 = 3 x 35 = 105 possibilities
2. 2 k + 3 b or 1 k + 4 b
3C2 * 7C3 = 3 x 35 = 105 possibilities
3. parcel 1 is equal to parcel 2 => only 105 possibilities at all
|
|
|
|
|
|
Director
Joined: 29 Oct 2004
Posts: 894
Followers: 2
Kudos [?]:
20
[0], given: 0
|
HongHu wrote: We need to select 5 from 10, with at least 1 kiwi.
1 kiwi
C(3,1)C(7,4)=3*7*6*5/6=105
And that's it. No need to do two kiwis since it is symmetric.
Can you explain a little more?
According to your result, it only answers this question:
In how many ways can John divide the 10 fruit between two parcels, if there has to be an equal total number of fruit in either parcel and so that there is only one kiwi in one parcel and two in the other.
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
It's the same. One in this bag means two in the other bag. And two in this bag means one in the other bag. Say you only have three fruits. I treat the following two case as the same outcome:
A has f1 and B has f2, f3
A has f2, f3 and B has f1
Will have to confirm with the OA though.
|
|
|
|
|
|
Director
Joined: 29 Oct 2004
Posts: 894
Followers: 2
Kudos [?]:
20
[0], given: 0
|
But that only applies when the bags are identical though. Right?
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
qhoc0010 wrote: But that only applies when the bags are identical though. Right?
Yes, applies to this case since we are making two parcels, nothing special with either of them.
|
|
|
|
|
|
Manager
Joined: 24 Jan 2005
Posts: 220
Location: Boston
Followers: 1
Kudos [?]:
3
[0], given: 0
|
7B + 3K in two parcels say I and II
No of ways :
(a) 1K in I and 1K in II == 3*2 = 6 ways
(b) 1K1B in I and 1K1B or 2K in II and vice versa = = 3*7*(2*6+2)*2=21(14)2=588 ways
Add 594 ways
Anirban
|
|
|
|
|
|
Manager
Joined: 24 Jan 2005
Posts: 220
Location: Boston
Followers: 1
Kudos [?]:
3
[0], given: 0
|
I assumed that the bananas and kiwis are not identical i mean in terms of shape size or color. So 7 distinct bananas and 3 distinct Kiwis.
Anirban
|
|
|
|
|
|
Senior Manager
Joined: 07 Oct 2003
Posts: 380
Location: Manhattan
Followers: 1
Kudos [?]:
2
[0], given: 0
|
HongHu wrote: We need to select 5 from 10, with at least 1 kiwi.
1 kiwi
C(3,1)C(7,4)=3*7*6*5/6=105
And that's it. No need to do two kiwis since it is symmetric.
Here's what I don't get. I see two possiblities:
A. 4 bananas and 1 kiwi (7c4 * 3c1) 105 possibilities
B. 3 bananas and 2 kiwi (7c3 * 3c2) 105 possibilities
total # of poss = 210
what am I missing? I understand that the two parcels are symmetric, but you still have different fruits in case one vs. case two.
|
|
|
|
|
|
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan
Followers: 1
Kudos [?]:
0
[0], given: 0
|
box1- 7C3 AND 3C2 OR 7C4 AND 3C1 = 105 + 105 = 210
box2- 7C4 AND 3C1 OR 7C3 AND 3C2 = 105 + 105 = 210
TOTAL=210+210 = 420 possibilities.
now where am i going wrong???????????
can anybody tell me plz
|
|
|
|
|
|
SVP
Joined: 30 Sep 2004
Posts: 1548
Location: Germany
Followers: 4
Kudos [?]:
15
[0], given: 0
|
faizaniftikhar83 wrote: box1- 7C3 AND 3C2 OR 7C4 AND 3C1 = 105 + 105 = 210
box2- 7C4 AND 3C1 OR 7C3 AND 3C2 = 105 + 105 = 210
TOTAL=210+210 = 420 possibilities.
now where am i going wrong???????????
can anybody tell me plz
parcel 1: kbbbb (case1) OR kkbbb(case2) = 105 p (you misused OR as AND)
parcel 2 : kkbbb(case1) OR kbbbb(case2) = 105 p (you misused OR as AND)
it is a combination problem that means it does not matter which parcel has the combination; it only matters the different way of combinations; in your approach you count the combination that appears in parcel 1 as well as the combination in parcel 2, but you should count it only once. maybe it helps 
|
|
|
|
|
|
Director
Joined: 19 Feb 2005
Posts: 508
Location: Milan Italy
Followers: 1
Kudos [?]:
2
[0], given: 0
|
What if it had been without the limitation of at least 1 kiwi per parcel?
|
|
|
|
|
|
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan
Followers: 1
Kudos [?]:
0
[0], given: 0
|
christoph wrote: faizaniftikhar83 wrote: box1- 7C3 AND 3C2 OR 7C4 AND 3C1 = 105 + 105 = 210
box2- 7C4 AND 3C1 OR 7C3 AND 3C2 = 105 + 105 = 210
TOTAL=210+210 = 420 possibilities.
now where am i going wrong???????????
can anybody tell me plz parcel 1: kbbbb (case1) OR kkbbb(case2) = 105 p (you misused OR as AND) parcel 2 : kkbbb(case1) OR kbbbb(case2) = 105 p (you misused OR as AND) it is a combination problem that means it does not matter which parcel has the combination; it only matters the different way of combinations; in your approach you count the combination that appears in parcel 1 as well as the combination in parcel 2, but you should count it only once. maybe it helps still...we get 210 ways....then howcome 105????
|
|
|
|
|
|
Manager
Joined: 01 Jan 2005
Posts: 169
Location: NJ
Followers: 1
Kudos [?]:
0
[0], given: 0
|
hmmm. earlier I calculated 210 but then I saw explainations and understood why it is 105.
|
|
|
|
|
|
Manager
Joined: 27 Jan 2005
Posts: 103
Location: San Jose,USA- India
Followers: 1
Kudos [?]:
1
[0], given: 0
|
ways to pick 5 fruits from 10 - ways to pick 5 friuts with all 3 kiwis - ways to pick 5 fruits with no kiwis
=10C5-8C3-7C5
=175
Note in both parcels there should be atleast 1 kiwi.Meaning you can not have 3 kiwis picked in the first parcel(second parcel will have zero kiwis), andyou can not have no kiwis at all.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
