How to solve this problem ?

[JHY]

Question :
Given six numbered cards with the number
3 4 5 6 7 8
Find the number of different numbers greater than 6500 that can be formed.


What I have tried is that
a) if the number in front is 6 , then the second digit must be 5 or 7 or 8, the third digit can be any number, the forth digit can be any number
so number of ways = 1 x 3 x 4 x 3
b) if the first number is 7 or 8, then the second digit can be any number, the third digit can be any number, the forth digit can be any number
so number of ways = 2 x 5 x 4 x 3
Therefore the total different numbers = 1 x 3 x 4 x 3 + 2 x 5 x 4 x 3 = 156

But the answer given is 132. I can't get this answer. What have I done wrongly?

 

[JeffM]

Thank you very much for showing your work.

Does the question restrict the numbers to be counted to numbers expressible in four digits? In fact, what EXACTLY does the problem say?

 

[blamocur]

Question :
Given six numbered cards with the number
3 4 5 6 7 8
Find the number of different numbers greater than 6500 that can be formed.


What I have tried is that
a) if the number in front is 6 , then the second digit must be 5 or 7 or 8, the third digit can be any number, the forth digit can be any number
so number of ways = 1 x 3 x 4 x 3
b) if the first number is 7 or 8, then the second digit can be any number, the third digit can be any number, the forth digit can be any number
so number of ways = 2 x 5 x 4 x 3
Therefore the total different numbers = 1 x 3 x 4 x 3 + 2 x 5 x 4 x 3 = 156

But the answer given is 132. I can't get this answer. What have I done wrongly?

I get the same answer when only 4-digit numbers are counted. But if one allows 5- and 6-digit answers then the numbers become even larger, so 132 does not look like a correct solution to me.

 

[blamocur]

Question :
Given six numbered cards with the number
3 4 5 6 7 8
Find the number of different numbers greater than 6500 that can be formed.


What I have tried is that
a) if the number in front is 6 , then the second digit must be 5 or 7 or 8, the third digit can be any number, the forth digit can be any number
so number of ways = 1 x 3 x 4 x 3
b) if the first number is 7 or 8, then the second digit can be any number, the third digit can be any number, the forth digit can be any number
so number of ways = 2 x 5 x 4 x 3
Therefore the total different numbers = 1 x 3 x 4 x 3 + 2 x 5 x 4 x 3 = 156

But the answer given is 132. I can't get this answer. What have I done wrongly?

BTW, one gets the answer of 132 if one replaces 6500 with 6800 and only counts 4-digit numbers.

 

[JHY]

Thank you very much for showing your work.

Does the question restrict the numbers to be counted to numbers expressible in four digits? In fact, what EXACTLY does the problem say?

"four digit number is to be formed by using four of these cards." I am sorry, I missed these part of the question while typing out the question when posting. Thank you for pointing it out to me.

So the number of different 4-digit numbers greater than 6500 that can be formed should be 156

and if 6500 is replaced by 6800, the answer is 132

Am I correct ?

 

[blamocur]

So the number of different 4-digit numbers greater than 6500 that can be formed should be 156

and if 6500 is replaced by 6800, the answer is 132

Am I correct ?

I believe you are correct.

 

[JeffM]

number greater than 6800 but under 7000 is 1 x 4 x 3 = 12

number greater than 7000 is 2 x 5 x 4 x 3 = 120

120 +12 = 132 looks good to me.

 

[Steven G]

One should note that saying given six numbered cards with the number 3 4 5 6 7 8 implies that no number can be used more than once. I hope that the OP noticed that.

 

[JHY]

Thank you very much for taking time to answer my question !