BDMO Regional Junior P8

For students of class 6-8 (age 12 to 14)
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Enthurelxyz
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BDMO Regional Junior P8

Unread post by Enthurelxyz » Sun Mar 28, 2021 9:22 am

How many subsets of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ contain $5$ consecutive numbers?
Last edited by tanmoy on Fri Apr 02, 2021 7:41 pm, edited 1 time in total.
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Mehrab4226
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Re: BDMO Regional Junior P8

Unread post by Mehrab4226 » Sun Mar 28, 2021 9:52 am

Enthurelxyz wrote:
Sun Mar 28, 2021 9:22 am
How many subsets of {1,2,3,4,5,6,7,8,9} contain $5$ consecutive numbers?
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Hese $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Similarly for case $3,4,5$ we get $2^2,2^1,2^0$ different $X$.
So total number of $X$ possible is $2^4+2^3+2^2+2^1+2^0=31$
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

ShurjoA
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Re: BDMO Regional Junior P8

Unread post by ShurjoA » Sun Mar 28, 2021 7:17 pm

User Mehrab4226 has already given the solution, though here I have shown a similar approach.

Firstly, let's look at how many sets there are with $5$ elements that satisfy the question. It is easy to see there are five sets, namely $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{4, 5, 6, 7, 8\}$ and $\{5, 6, 7, 8, 9\}.$

Secondly, let's look at how many sets there are with $6$ elements that satisfy the question. It could be helpful to use the previous sets. Let's start with the set $\{1, 2, 3, 4, 5\}.$ To make this a set with $6$ elements, we will need to choose a number from the remaining four numbers. We can write that as $^4C_1.$ Next, let's look at the set $\{2, 3, 4, 5, 6\}.$ Here, we have 3 remaining numbers. We can also write that as $^3C_1.$ Doing this, we get that the number of sets with $6$ elements is $^4C_1 + ^3C_1 + ^2C_1 + ^1C_1 = 4+3+2+1 = 10.$

Finally, similar to the sets with six elements, we get that the number of sets with $7$ elements is $^4C_2 + ^3C_2 + ^2C_2 = 6+3+1 = 10,$ the number of sets with $8$ elements is $^4C_2 + ^3C_2 = 4+1 = 5,$ and the number of sets with $9$ elements is $^4C_4 = 1$. So the total number of sets/subsets is $5+10+10+5+1 = 31.$ (Answer)

Shurjo :)

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Enthurelxyz
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Re: BDMO Regional Junior P8

Unread post by Enthurelxyz » Tue Mar 30, 2021 12:50 pm

Mehrab4226 wrote:
Sun Mar 28, 2021 9:52 am
Enthurelxyz wrote:
Sun Mar 28, 2021 9:22 am
How many subsets of {1,2,3,4,5,6,7,8,9} contain $5$ consecutive numbers?
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Hese $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Similarly for case $3,4,5$ we get $2^2,2^1,2^0$ different $X$.
So total number of $X$ possible is $2^4+2^3+2^2+2^1+2^0=31$
Have you counted the set $\{1,3,4,5,6,7\}$?
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Mehrab4226
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Re: BDMO Regional Junior P8

Unread post by Mehrab4226 » Tue Mar 30, 2021 1:50 pm

Enthurelxyz wrote:
Tue Mar 30, 2021 12:50 pm
Mehrab4226 wrote:
Sun Mar 28, 2021 9:52 am
Enthurelxyz wrote:
Sun Mar 28, 2021 9:22 am
How many subsets of {1,2,3,4,5,6,7,8,9} contain $5$ consecutive numbers?
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Hese $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Similarly for case $3,4,5$ we get $2^2,2^1,2^0$ different $X$.
So total number of $X$ possible is $2^4+2^3+2^2+2^1+2^0=31$
Have you counted the set $\{1,3,4,5,6,7\}$?
Thank you for pointing that out.
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Here $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Case:3
Here $X$ are of this kind:$\{3,4,5,6,7\}\cup \{\text{Any subset of}\{1,8,9\}$
So total number of $X$ of this kind is $2^3$
Case:4
Here $X$ are of this kind:$\{4,5,6,7,8\}\cup \{\text{Any subset of}\{1,2,9\}$
So total number of $X$ of this kind is $2^3$
Case:5
Here $X$ are of kind:$\{5,6,7,8,9\}\cup\{\text{Any subset of}\{1,2,3\}$
so total number of $X$ of this kind is $2^3$
So the total number of $X$ we got is $2^4+4\times 2^3=48$(ans)
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

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Enthurelxyz
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Re: BDMO Regional Junior P8

Unread post by Enthurelxyz » Tue Mar 30, 2021 2:32 pm

Mehrab4226 wrote:
Tue Mar 30, 2021 1:50 pm
Enthurelxyz wrote:
Tue Mar 30, 2021 12:50 pm
Mehrab4226 wrote:
Sun Mar 28, 2021 9:52 am
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Hese $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Similarly for case $3,4,5$ we get $2^2,2^1,2^0$ different $X$.
So total number of $X$ possible is $2^4+2^3+2^2+2^1+2^0=31$
Have you counted the set $\{1,3,4,5,6,7\}$?
Thank you for pointing that out.
Let us denote that kind of subsets as $X$
At first, we look at how many sets of $5$ consecutive numbers are there.
  • 1.$\{1,2,3,4,5\}$
    2.$\{2,3,4,5,6\}$
    3.$\{3,4,5,6,7\}$
    4.$\{4,5,6,7,8\}$
    5.$\{5,6,7,8,9\}$
Now we will divide our work in $5$ cases. Case $1$ represents when number$ 1$. of the list is within the subsets we are looking for.
Case:1
Here $X$ are of this kind:$\{1,2,3,4,5\}\cup \{\text{Any subset of}\{6,7,8,9\}\}$
So total number of $X$ of this kind is $2^4$
Case:2
Here $X$ are of this kind: $\{2,3,4,5,6\}\cup \{\text{Any subset of}\{7,8,9\}\}$
So the total number of $X$ of this kind $2^3$
Case:3
Here $X$ are of this kind:$\{3,4,5,6,7\}\cup \{\text{Any subset of}\{1,8,9\}$
So total number of $X$ of this kind is $2^3$
Case:4
Here $X$ are of this kind:$\{4,5,6,7,8\}\cup \{\text{Any subset of}\{1,2,9\}$
So total number of $X$ of this kind is $2^3$
Case:5
Here $X$ are of kind:$\{5,6,7,8,9\}\cup\{\text{Any subset of}\{1,2,3\}$
so total number of $X$ of this kind is $2^3$
So the total number of $X$ we got is $2^4+4\times 2^3=48$(ans)
Nice and Thanks. :D
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Enthurelxyz
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Re: BDMO Regional Junior P8

Unread post by Enthurelxyz » Tue Mar 30, 2021 2:33 pm

ShurjoA wrote:
Sun Mar 28, 2021 7:17 pm
User Mehrab4226 has already given the solution, though here I have shown a similar approach.

Firstly, let's look at how many sets there are with $5$ elements that satisfy the question. It is easy to see there are five sets, namely $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{4, 5, 6, 7, 8\}$ and $\{5, 6, 7, 8, 9\}.$

Secondly, let's look at how many sets there are with $6$ elements that satisfy the question. It could be helpful to use the previous sets. Let's start with the set $\{1, 2, 3, 4, 5\}.$ To make this a set with $6$ elements, we will need to choose a number from the remaining four numbers. We can write that as $^4C_1.$ Next, let's look at the set $\{2, 3, 4, 5, 6\}.$ Here, we have 3 remaining numbers. We can also write that as $^3C_1.$ Doing this, we get that the number of sets with $6$ elements is $^4C_1 + ^3C_1 + ^2C_1 + ^1C_1 = 4+3+2+1 = 10.$

Finally, similar to the sets with six elements, we get that the number of sets with $7$ elements is $^4C_2 + ^3C_2 + ^2C_2 = 6+3+1 = 10,$ the number of sets with $8$ elements is $^4C_2 + ^3C_2 = 4+1 = 5,$ and the number of sets with $9$ elements is $^4C_4 = 1$. So the total number of sets/subsets is $5+10+10+5+1 = 31.$ (Answer)

Shurjo :)
I think 31 is not the right answer.
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ShurjoA
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Re: BDMO Regional Junior P8

Unread post by ShurjoA » Wed Mar 31, 2021 12:11 pm

Enthurelxyz wrote:
Tue Mar 30, 2021 2:33 pm
ShurjoA wrote:
Sun Mar 28, 2021 7:17 pm
User Mehrab4226 has already given the solution, though here I have shown a similar approach.

Firstly, let's look at how many sets there are with $5$ elements that satisfy the question. It is easy to see there are five sets, namely $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{4, 5, 6, 7, 8\}$ and $\{5, 6, 7, 8, 9\}.$

Secondly, let's look at how many sets there are with $6$ elements that satisfy the question. It could be helpful to use the previous sets. Let's start with the set $\{1, 2, 3, 4, 5\}.$ To make this a set with $6$ elements, we will need to choose a number from the remaining four numbers. We can write that as $^4C_1.$ Next, let's look at the set $\{2, 3, 4, 5, 6\}.$ Here, we have 3 remaining numbers. We can also write that as $^3C_1.$ Doing this, we get that the number of sets with $6$ elements is $^4C_1 + ^3C_1 + ^2C_1 + ^1C_1 = 4+3+2+1 = 10.$

Finally, similar to the sets with six elements, we get that the number of sets with $7$ elements is $^4C_2 + ^3C_2 + ^2C_2 = 6+3+1 = 10,$ the number of sets with $8$ elements is $^4C_2 + ^3C_2 = 4+1 = 5,$ and the number of sets with $9$ elements is $^4C_4 = 1$. So the total number of sets/subsets is $5+10+10+5+1 = 31.$ (Answer)

Shurjo :)
I think 31 is not the right answer.
Thanks. I saw the other solution. It's correct. You can check that one.
2+2 is 4, -1 that's 3 quick maths!

M F Ruhan
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Re: BDMO Regional Junior P8

Unread post by M F Ruhan » 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


Another possible solution is:
Solution 2:
If we consider these 5 main subsets
{1,2,3,4,5} {2,3,4,5,6} {3,4,5,6,7} {4,5,6,7,8} {5,6,7,8,9}
then total such subset should be 5* 2^(9-5) =80

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Mehrab4226
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Re: BDMO Regional Junior P8

Unread post by Mehrab4226 » 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.
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

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