## BDMO Regional Junior P8

- Enthurelxyz
**Posts:**17**Joined:**Sat Dec 05, 2020 10:45 pm**Location:**Bangladesh-
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### 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

and miles to go before we sleep

- Mehrab4226
**Posts:**214**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BDMO Regional Junior P8

Enthurelxyz wrote: ↑Sun Mar 28, 2021 9:22 amHow 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

### 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

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**Location:**Bangladesh-
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### Re: BDMO Regional Junior P8

Have you counted the set $\{1,3,4,5,6,7\}$?Mehrab4226 wrote: ↑Sun Mar 28, 2021 9:52 amEnthurelxyz wrote: ↑Sun Mar 28, 2021 9:22 amHow many subsets of {1,2,3,4,5,6,7,8,9} contain $5$ consecutive numbers?

and miles to go before we sleep

and miles to go before we sleep

and miles to go before we sleep

- Mehrab4226
**Posts:**214**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BDMO Regional Junior P8

Thank you for pointing that out.Enthurelxyz wrote: ↑Tue Mar 30, 2021 12:50 pmHave you counted the set $\{1,3,4,5,6,7\}$?Mehrab4226 wrote: ↑Sun Mar 28, 2021 9:52 amEnthurelxyz wrote: ↑Sun Mar 28, 2021 9:22 amHow 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

- Enthurelxyz
**Posts:**17**Joined:**Sat Dec 05, 2020 10:45 pm**Location:**Bangladesh-
**Contact:**

### Re: BDMO Regional Junior P8

Nice and Thanks.Mehrab4226 wrote: ↑Tue Mar 30, 2021 1:50 pmThank you for pointing that out.

and miles to go before we sleep

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**Location:**Bangladesh-
**Contact:**

### Re: BDMO Regional Junior P8

I think 31 is not the right answer.ShurjoA wrote: ↑Sun Mar 28, 2021 7:17 pmUser 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

and miles to go before we sleep

and miles to go before we sleep

and miles to go before we sleep

### Re: BDMO Regional Junior P8

Thanks. I saw the other solution. It's correct. You can check that one.Enthurelxyz wrote: ↑Tue Mar 30, 2021 2:33 pmI think 31 is not the right answer.ShurjoA wrote: ↑Sun Mar 28, 2021 7:17 pmUser 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

2+2 is 4, -1 that's 3 quick maths!

### 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

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

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:**214**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BDMO Regional Junior P8

we 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

*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