## BDMO Regional Junior P8

For students of class 6-8 (age 12 to 14)
Enthurelxyz
Posts: 17
Joined: Sat Dec 05, 2020 10:45 pm
Contact:

### BDMO Regional Junior P8

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.
Reason: The author didn't use Latex.
and miles to go before we sleep
and miles to go before we sleep

Mehrab4226
Posts: 208
Joined: Sat Jan 11, 2020 1:38 pm

### Re: BDMO Regional Junior P8

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?
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
Posts: 8
Joined: Thu Dec 03, 2020 12:02 pm

### Re: BDMO Regional Junior P8

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

Enthurelxyz
Posts: 17
Joined: Sat Dec 05, 2020 10:45 pm
Contact:

### Re: BDMO Regional Junior P8

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?
Have you counted the set $\{1,3,4,5,6,7\}$?
and miles to go before we sleep
and miles to go before we sleep

Mehrab4226
Posts: 208
Joined: Sat Jan 11, 2020 1:38 pm

### Re: BDMO Regional Junior P8

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?
Have you counted the set $\{1,3,4,5,6,7\}$?
Thank you for pointing that out.
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

Enthurelxyz
Posts: 17
Joined: Sat Dec 05, 2020 10:45 pm
Contact:

### Re: BDMO Regional Junior P8

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
Have you counted the set $\{1,3,4,5,6,7\}$?
Thank you for pointing that out.
Nice and Thanks.
and miles to go before we sleep
and miles to go before we sleep

Enthurelxyz
Posts: 17
Joined: Sat Dec 05, 2020 10:45 pm
Contact:

### Re: BDMO Regional Junior P8

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.
and miles to go before we sleep
and miles to go before we sleep

ShurjoA
Posts: 8
Joined: Thu Dec 03, 2020 12:02 pm

### Re: BDMO Regional Junior P8

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

### Re: BDMO Regional Junior P8

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

Mehrab4226
Posts: 208
Joined: Sat Jan 11, 2020 1:38 pm

### Re: BDMO Regional Junior P8

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