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number of trees
Source: Beat The GMAT — Problem Solving |
-
indiantiger
- Master | Next Rank: 500 Posts
- Posts: 362
- Thanked: 26 times
I am getting the ans as 10P3*3^7..this is not in option.
logic is he has to plant 3 trees in 3 day(one each) we can select and arrange 3 trees from 10 in 10P3 ways
for the remaining 7 trees each can be planted in any of the 3 days so to tal 3*3*3*3*3*3*3=3^7
So ans 10P3*3^7
Can any body please check what is wrong with this approach?
logic is he has to plant 3 trees in 3 day(one each) we can select and arrange 3 trees from 10 in 10P3 ways
for the remaining 7 trees each can be planted in any of the 3 days so to tal 3*3*3*3*3*3*3=3^7
So ans 10P3*3^7
Can any body please check what is wrong with this approach?
"If you don't know where you are going, any road will get you there."
Lewis Carroll
Lewis Carroll
Do not know how to approach this question but Process Of Elimination is good friend!
I can readily make more than 12 combinations hence the answer is 36 ........
I can readily make more than 12 combinations hence the answer is 36 ........
hi guys
sorry for posting 2 identical answers, it was a typo and i corrected it
the answer to the problem -36
but i did not see any smart way to solve,
i actually count all number of possible outcomes ,after some day the combinations start to repeat
but for sure it is not the way to follow as prone to mistake and time consuming
one smart math student advise me to look at the problem as:
in how many way we can fill 3 boxes with 10 balls so that every box contain at least 1 box but any way not clear
i do not think this problem too hard the only we need is good approach
sorry for posting 2 identical answers, it was a typo and i corrected it
the answer to the problem -36
but i did not see any smart way to solve,
i actually count all number of possible outcomes ,after some day the combinations start to repeat
but for sure it is not the way to follow as prone to mistake and time consuming
one smart math student advise me to look at the problem as:
in how many way we can fill 3 boxes with 10 balls so that every box contain at least 1 box but any way not clear
i do not think this problem too hard the only we need is good approach
I do not know if this is the most efficient but it is an approach nonetheless.
Say he plants 1 tree in day one, now on day 2 and 3 he can plant 9 tress in 8 ways - 8,1;7,2;3,6;4,5;5,4;6,3;7,2;8,1
Now if he plants 2 trees in day one, now on day 2 and 3 he can plant 8 trees in 7 ways: 1,7;2,6;3,5;4,4;5,3;6,2;7,1
Thus, we can go all the way to 8 trees in one day and adding up ways of doing this is: 8+7+6+5+4+3+2+1 = 36 ways.
Hence, my answer is B
Say he plants 1 tree in day one, now on day 2 and 3 he can plant 9 tress in 8 ways - 8,1;7,2;3,6;4,5;5,4;6,3;7,2;8,1
Now if he plants 2 trees in day one, now on day 2 and 3 he can plant 8 trees in 7 ways: 1,7;2,6;3,5;4,4;5,3;6,2;7,1
Thus, we can go all the way to 8 trees in one day and adding up ways of doing this is: 8+7+6+5+4+3+2+1 = 36 ways.
Hence, my answer is B
One tree has to be planted in each of the three days
Left with 10 -3 = 7 which can be planted in any of the three days.
is this is possible in 7 C 3 ways ??
Left with 10 -3 = 7 which can be planted in any of the three days.
is this is possible in 7 C 3 ways ??
hi
i was told how to solve this and want to share the way with others
the first step is the same as mentioned kstv
at first plant 1 tree in 1 day, and left with 10-3=7 trees
then let us write in the line 7 trees designate them with 0 and days as (___)
0 0 0 0 0 0 0 ___ ____
it means that 1 the first day gardener will plant 7 trees and nothing in remaining 2 days he plants nothing
or it may look like
0 0 ___ 0 0____ 0 0 0
1 st day 2 trees, 2-d day 2 trees and third day 3 trees
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
i was told how to solve this and want to share the way with others
the first step is the same as mentioned kstv
at first plant 1 tree in 1 day, and left with 10-3=7 trees
then let us write in the line 7 trees designate them with 0 and days as (___)
0 0 0 0 0 0 0 ___ ____
it means that 1 the first day gardener will plant 7 trees and nothing in remaining 2 days he plants nothing
or it may look like
0 0 ___ 0 0____ 0 0 0
1 st day 2 trees, 2-d day 2 trees and third day 3 trees
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
ID the permutation of three positive integers that when added = 10
I. 1 + 2 + 7
II. 1 + 3 + 6
III. 1 + 4 + 5
IV. 2 + 3 + 5
V. 2 + 4 + 4
VI. 3 + 3 + 4
Total 6 sets of 3
Permutation of sets of 3 = 3! = 3*2*1 = 6
Therefore, 6 * 6 = 36
I. 1 + 2 + 7
II. 1 + 3 + 6
III. 1 + 4 + 5
IV. 2 + 3 + 5
V. 2 + 4 + 4
VI. 3 + 3 + 4
Total 6 sets of 3
Permutation of sets of 3 = 3! = 3*2*1 = 6
Therefore, 6 * 6 = 36
-
jeffedwards
- Master | Next Rank: 500 Posts
- Posts: 109
- Thanked: 12 times
- GMAT Score:720
will you explain to me how you got 9C2? just trying to understandclock60 wrote:hi
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
(k+m-1)C(m)=(k+m-1)!/(m!(k-1)!)jeffedwards wrote:will you explain to me how you got 9C2? just trying to understandclock60 wrote:hi
we can vary it
and the total number of outcomes will looks like
9!/(7!*2!)=36
i am not sure but it is combination with repetition
It's not mentioned, but all of 10 trees are thought to be identical (otherwise the number of ways obviously is greater than 48).
Let us suppose there is a box with two partitions, so we have three distinct parts. It may look like that:
\ **** | * | ***** /
where '\' and '/' are box' immovable walls, '*' is one of 10 trees, and '|' is a partition. Let us think that the gardener plants trees from the first part on the first day, trees from the second part on the second day, etc.
Now we have to obtain the number of ways to place 10 trees in 3 sub-boxes. But instead of moving trees from one part to another, we'll move the partitions in all ways possible. There're 9 possible places for 2 partitions, hence the overall number of ways to place them is a binomial coefficient 9C2 = (8*9) / 2! = 8*9 / 2 = 4*9 = 36.
Answer B.
Let us suppose there is a box with two partitions, so we have three distinct parts. It may look like that:
\ **** | * | ***** /
where '\' and '/' are box' immovable walls, '*' is one of 10 trees, and '|' is a partition. Let us think that the gardener plants trees from the first part on the first day, trees from the second part on the second day, etc.
Now we have to obtain the number of ways to place 10 trees in 3 sub-boxes. But instead of moving trees from one part to another, we'll move the partitions in all ways possible. There're 9 possible places for 2 partitions, hence the overall number of ways to place them is a binomial coefficient 9C2 = (8*9) / 2! = 8*9 / 2 = 4*9 = 36.
Answer B.
