BDMO Regional Junior P8

For students of class 6-8 (age 12 to 14)
M F Ruhan
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Re: BDMO Regional Junior P8

Unread post by M F Ruhan » Sat Apr 03, 2021 2:00 pm

Mehrab4226 wrote:
Fri Apr 02, 2021 9:32 pm
M F Ruhan wrote:
Fri Apr 02, 2021 2:12 pm
Solution 1:
At first consider 1 such simple subset {1,2,3,4,5}
Now we can create any combination this subset by using numbers 6,7,8,9
for example: {1,2,3,4,5,8,9}
Total number of combination for this subset should be 4C4+4C3+4C2+4C1=15

As we can create total 5 simple subsets. So the total number of subsets should be 5*15+5=80
we cannot create combinations of 4 numbers on all those 5 subsets. And the answer should be 48.
Why can't we create combinations of 4 numbers on all those subsets? Would you please explain :?:

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Mehrab4226
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Location:Dhaka, Bangladesh

Re: BDMO Regional Junior P8

Unread post by Mehrab4226 » Sat Apr 03, 2021 7:06 pm

M F Ruhan wrote:
Sat Apr 03, 2021 2:00 pm
Mehrab4226 wrote:
Fri Apr 02, 2021 9:32 pm
M F Ruhan wrote:
Fri Apr 02, 2021 2:12 pm
Solution 1:
At first consider 1 such simple subset {1,2,3,4,5}
Now we can create any combination this subset by using numbers 6,7,8,9
for example: {1,2,3,4,5,8,9}
Total number of combination for this subset should be 4C4+4C3+4C2+4C1=15

As we can create total 5 simple subsets. So the total number of subsets should be 5*15+5=80
we cannot create combinations of 4 numbers on all those 5 subsets. And the answer should be 48.
Why can't we create combinations of 4 numbers on all those subsets? Would you please explain :?:
we can creat what you are saying in $\{1,2,3,4,5\}$ subset. But we cannot do the same for $\{3,4,5,6,7\}$. We cannot make combinations with $\{6,7,8,9\}$ because 6,7 is already on the main subset.
Last edited by Mehrab4226 on Sat Apr 03, 2021 9:03 pm, edited 1 time in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

M F Ruhan
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Joined:Fri Apr 02, 2021 1:42 pm
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Re: BDMO Regional Junior P8

Unread post by M F Ruhan » Sat Apr 03, 2021 8:20 pm

Mehrab4226 wrote:
Sat Apr 03, 2021 7:06 pm
M F Ruhan wrote:
Sat Apr 03, 2021 2:00 pm
Mehrab4226 wrote:
Fri Apr 02, 2021 9:32 pm

we cannot create combinations of 4 numbers on all those 5 subsets. And the answer should be 48.
Why can't we create combinations of 4 numbers on all those subsets? Would you please explain :?:
we can creat what you are saying in $\{1,2,3,4,5\} subset. But we cannot do the same for $\{3,4,5,6,7\}$. We cannot make combinations with $\{6,7,8,9\}$ because 6,7 is already on the main subset.
Yes, we can't, but as result we can create new combinations using 1,2 . As you can see for each main subset always extra 4 numbers are remaining . For example-1,7,8,9 for{2,3,4,5,6}
1,2,8,9 for {3,4,5,6,7} and so on for the other 3.

Does your solution contain subsets like {2,4,5,6,7,8} or {1,3,4,5,6,7,8} :?:

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Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: BDMO Regional Junior P8

Unread post by Mehrab4226 » Sat Apr 03, 2021 9:01 pm

M F Ruhan wrote:
Sat Apr 03, 2021 8:20 pm
Mehrab4226 wrote:
Sat Apr 03, 2021 7:06 pm
M F Ruhan wrote:
Sat Apr 03, 2021 2:00 pm


Why can't we create combinations of 4 numbers on all those subsets? Would you please explain :?:
we can creat what you are saying in $\{1,2,3,4,5\} subset. But we cannot do the same for $\{3,4,5,6,7\}$. We cannot make combinations with $\{6,7,8,9\}$ because 6,7 is already on the main subset.
Yes, we can't, but as result we can create new combinations using 1,2 . As you can see for each main subset always extra 4 numbers are remaining . For example-1,7,8,9 for{2,3,4,5,6}
1,2,8,9 for {3,4,5,6,7} and so on for the other 3.

Does your solution contain subsets like {2,4,5,6,7,8} or {1,3,4,5,6,7,8} :?:
Then what you are doing is counting the same thing multiple times.
You once took $\{1,2,3,4,5\}$ and combined it with $6$ and got $\{1,2,3,4,5,6\}$ to get $\{1,2,3,4,5,6\}$. Again you took $\{2,3,4,5,6\}$ combined it with $1$ and got $\{1,2,3,4,5,6\}$. Which is the same set, but you are counting it multiple times. Read my 2nd solution, it avoids the error.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

M F Ruhan
Posts:13
Joined:Fri Apr 02, 2021 1:42 pm
Contact:

Re: BDMO Regional Junior P8

Unread post by M F Ruhan » Sat Apr 03, 2021 11:39 pm

Mehrab4226 wrote:
Sat Apr 03, 2021 9:01 pm
M F Ruhan wrote:
Sat Apr 03, 2021 8:20 pm
Mehrab4226 wrote:
Sat Apr 03, 2021 7:06 pm

we can creat what you are saying in $\{1,2,3,4,5\} subset. But we cannot do the same for $\{3,4,5,6,7\}$. We cannot make combinations with $\{6,7,8,9\}$ because 6,7 is already on the main subset.
Yes, we can't, but as result we can create new combinations using 1,2 . As you can see for each main subset always extra 4 numbers are remaining . For example-1,7,8,9 for{2,3,4,5,6}
1,2,8,9 for {3,4,5,6,7} and so on for the other 3.

Does your solution contain subsets like {2,4,5,6,7,8} or {1,3,4,5,6,7,8} :?:
Then what you are doing is counting the same thing multiple times.
You once took $\{1,2,3,4,5\}$ and combined it with $6$ and got $\{1,2,3,4,5,6\}$ to get $\{1,2,3,4,5,6\}$. Again you took $\{2,3,4,5,6\}$ combined it with $1$ and got $\{1,2,3,4,5,6\}$. Which is the same set, but you are counting it multiple times. Read my 2nd solution, it avoids the error.
I finally got it. Thank you very much for correcting me. :D

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