How Many Unique 7-Digit Numbers Can You Form Under Different Conditions?

[jinx007]

Please i need help i am not that good in probability and permutation.

The digits of the number 1,2,2,3,6,7,8 can be read to give many 7-digits numbers. Find how many different 7-digit numbers can be made if

1/ There is no restriction on the order of the digits.
2/ The digits 1,3,7(in any order) are next to each other
3/ These 7-digits numbers are even

1/ if no restriction so 7 ! = 5040 ways

2/ so here i treat 1,3,7 as a block and is equal to 1 ! and i add to the remaining and i get 4 ! + 1 ! = 5!

And my answer = 5! x 3

But i am not sure...please check it..and tell me if there is any error

3/ I treat the last block as even so (2,2,6,8)...> 4!

my answer 6 x 5 x 4 x 3 x 2 x 1 x 4! = 17280 ways


ALL WHAT I HAVE DONE, I AM NOT SURE..PLEASE CHECK IT...I HAVE MANY DOUBT...THANKS IN ADVANCE

 

[lanedance]

are you sure about 1)...? note the fact you have two 2's that are, for all intents & purposes, indistinguishable...

 

[lanedance]

the point made above will feed into 2 & 3

for 2)
note there are 3! different ways to arrange 1,3 & 7
and I'm not sure what you mean by 1! + 4! = 5!, as this is not true...
but reading between the lines, i think you've got the right idea

 

[jinx007]

the point made above will feed into 2 & 3

for 2)
note there are 3! different ways to arrange 1,3 & 7
and I'm not sure what you mean by 1! + 4! = 5!, as this is not true...
but reading between the lines, i think you've got the right idea

Ohhh for the first part i have some doubt..because of the two 2's...don't you know how to proceed..??

for the second part i make use of blocks technique that is the 1,3 and 7 acting as 1 block(so 1 !) and the remaining as 4 blocks (4 !) so i add them and multiply by 3 ( 3 because i think there are 3 ways) I think i must multiply by 3!

so the answer must be like that 5! x 3!

 

[lanedance]

yeah so when there is indistinguishabilty, just count them as if they were distinguishable, then divide by the number of different ways you could arrange the indistinguishable items.

that will acount for the repeated arrangements

 

[utkarsh009]

Please i need help i am not that good in probability and permutation.

The digits of the number 1,2,2,3,6,7,8 can be read to give many 7-digits numbers. Find how many different 7-digit numbers can be made if

1/ There is no restriction on the order of the digits.
2/ The digits 1,3,7(in any order) are next to each other
3/ These 7-digits numbers are even

1/ if no restriction so 7 ! = 5040 ways

2/ so here i treat 1,3,7 as a block and is equal to 1 ! and i add to the remaining and i get 4 ! + 1 ! = 5!

And my answer = 5! x 3

But i am not sure...please check it..and tell me if there is any error

3/ I treat the last block as even so (2,2,6,8)...> 4!

my answer 6 x 5 x 4 x 3 x 2 x 1 x 4! = 17280 ways


ALL WHAT I HAVE DONE, I AM NOT SURE..PLEASE CHECK IT...I HAVE MANY DOUBT...THANKS IN ADVANCE

if you have the answers, then please post them because all that you have done seem to be incorrect. your answers will ensure whether i am correct or not. if you don't have them then please post a message that you don't have them. then i will post my way of solving all the problems you have posted. thank you!

 

[utkarsh009]

here is my way of solving the problem .
1. in the first case 2 is repeating itself 2 times (there are two 2's ) therefore the number of different 7-digit numbers that can be formed are : 7!/2! = 5040/2 = 2520
2. as for the second case the answer is : 1,3,7 can be arranged in 3! ways, 2,2,6,8 can be arranged in 4!/2! ways i.e. 12 ways, these two blocks can be arranged in 2! ways. therefore the final answer is: 3!x12x2! = 6x12x2 = 144 ways
3. as for the third case the answer is : an even number must have even numbers in its unit place. therefore the number must end with either 2, 6 or 8. numbers formed by digits ending in 8 are: 6!/2! = 360 , the numbers formed by digits ending in 2 are: 6! = 720, the numbers ending in 6 are: 6!/2! = 360. therefore the total number of even 7-digit numbers formed are : 360+720+360 = 1440.
that was all i had. just match them with your answers and tell me if they are correct or not.

 

[lanedance]

hey utkarsh, the general idea is to help people through the problem rather than doing the whole thing for them ;)

here is my way of solving the problem .
1. in the first case 2 is repeating itself 2 times (there are two 2's ) therefore the number of different 7-digit numbers that can be formed are : 7!/2! = 5040/2 = 2520

this is ok

2. as for the second case the answer is : 1,3,7 can be arranged in 3! ways, 2,2,6,8 can be arranged in 4!/2! ways i.e. 12 ways, these two blocks can be arranged in 2! ways. therefore the final answer is: 3!x12x2! = 6x12x2 = 144 ways

i don't think this is quite right, as was originally done consider the arrangement of the 5 items, the block (137) & 2, 2, 6, 8 which can be arranged in 5!/2! different ways

then account for the 3! ways to arrange 137

3. as for the third case the answer is : an even number must have even numbers in its unit place. therefore the number must end with either 2, 6 or 8. numbers formed by digits ending in 8 are: 6!/2! = 360 , the numbers formed by digits ending in 2 are: 6! = 720, the numbers ending in 6 are: 6!/2! = 360. therefore the total number of even 7-digit numbers formed are : 360+720+360 = 1440.
that was all i had. just match them with your answers and tell me if they are correct or not.

this looks ok

 

[utkarsh009]

yes lanedance you are absolutely right. the second one was a silly mistake. thank you very much for pointing out my mistake. after all i am just a student of 9th standard. i often make mistake. and from now i will keep this in mind not to solve the problems just help. thank you again!

 

[jinx007]

here is my way of solving the problem .
1. in the first case 2 is repeating itself 2 times (there are two 2's ) therefore the number of different 7-digit numbers that can be formed are : 7!/2! = 5040/2 = 2520
2. as for the second case the answer is : 1,3,7 can be arranged in 3! ways, 2,2,6,8 can be arranged in 4!/2! ways i.e. 12 ways, these two blocks can be arranged in 2! ways. therefore the final answer is: 3!x12x2! = 6x12x2 = 144 ways
3. as for the third case the answer is : an even number must have even numbers in its unit place. therefore the number must end with either 2, 6 or 8. numbers formed by digits ending in 8 are: 6!/2! = 360 , the numbers formed by digits ending in 2 are: 6! = 720, the numbers ending in 6 are: 6!/2! = 360. therefore the total number of even 7-digit numbers formed are : 360+720+360 = 1440.
that was all i had. just match them with your answers and tell me if they are correct or not.

Ohhh i don't have the answer...but it seem your work is correct...thank

 

[utkarsh009]

Ohhh i don't have the answer...but it seem your work is correct...thank

thank you! but the second one has been done correctly by lanedance, just have a look.

 

[utkarsh009]

Ohhh i don't have the answer...but it seem your work is correct...thank

thank you! but the second one has been done correctly by lanedance. in that one the second block which i mentioned is not a block so the correct answer is that of lanedance. check that out

 

[lanedance]

yes lanedance you are absolutely right. the second one was a silly mistake. thank you very much for pointing out my mistake. after all i am just a student of 9th standard. i often make mistake. and from now i will keep this in mind not to solve the problems just help. thank you again!

no worries, it can be tempting to solve an interesting problem & i still do it sometimes though the idea is to help the OP get the most out of the problem