Why can't we create combinations of 4 numbers on all those subsets? Would you please explainMehrab4226 wrote: ↑Fri Apr 02, 2021 9:32 pmwe cannot create combinations of 4 numbers on all those 5 subsets. And the answer should be 48.M F Ruhan wrote: ↑Fri Apr 02, 2021 2:12 pmSolution 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
BDMO Regional Junior P8
- Mehrab4226
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Re: BDMO Regional Junior P8
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.M F Ruhan wrote: ↑Sat Apr 03, 2021 2:00 pmWhy can't we create combinations of 4 numbers on all those subsets? Would you please explainMehrab4226 wrote: ↑Fri Apr 02, 2021 9:32 pmwe cannot create combinations of 4 numbers on all those 5 subsets. And the answer should be 48.M F Ruhan wrote: ↑Fri Apr 02, 2021 2:12 pmSolution 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
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
-Henri Poincaré
Re: BDMO Regional Junior P8
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}Mehrab4226 wrote: ↑Sat Apr 03, 2021 7:06 pmwe 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.M F Ruhan wrote: ↑Sat Apr 03, 2021 2:00 pmWhy can't we create combinations of 4 numbers on all those subsets? Would you please explainMehrab4226 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.
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}
- Mehrab4226
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Re: BDMO Regional Junior P8
Then what you are doing is counting the same thing multiple times.M F Ruhan wrote: ↑Sat Apr 03, 2021 8:20 pmYes, 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}Mehrab4226 wrote: ↑Sat Apr 03, 2021 7:06 pmwe 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.
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}
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
-Henri Poincaré
Re: BDMO Regional Junior P8
I finally got it. Thank you very much for correcting me.Mehrab4226 wrote: ↑Sat Apr 03, 2021 9:01 pmThen what you are doing is counting the same thing multiple times.M F Ruhan wrote: ↑Sat Apr 03, 2021 8:20 pmYes, 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}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.
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}
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.