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 $2^4+2^3+2^2+2^1+2^0=31$
Re: BDMO Regional Junior P8
Posted: Sun Mar 28, 2021 7:17 pm
by ShurjoA
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)
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 $2^4+2^3+2^2+2^1+2^0=31$
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.
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)
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.
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)
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)
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.
Re: BDMO Regional Junior P8
Posted: Fri Apr 02, 2021 2:12 pm
by M F Ruhan
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
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.