- #1

Jurij

- 14

- 1

We have 5 digit. How many 7-digit numbers can we create that has at least 2 different digit?

Jurij

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- Thread starter Jurij
- Start date

- #1

Jurij

- 14

- 1

We have 5 digit. How many 7-digit numbers can we create that has at least 2 different digit?

Jurij

- #2

neurocomp2003

- 1,366

- 3

whats the problem?

- #3

Jurij

- 14

- 1

I don't know how to solve the task.

Could anybody help me?

Could anybody help me?

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- #4

Jurij

- 14

- 1

I think that the answer is 5^6=15625 but i don't know how to prove it. Could you help me?

- #5

mukundpa

Homework Helper

- 524

- 3

I think this problem is related to **permutation and combination** in maths.

- #6

David

Science Advisor

- 41

- 0

As I understand it, you are ask to create a seven-digit number where each digit comes from a set of five choices, and there must be at least two different digits represented within the seven. If that is the right interpretation of the question, there are a number of ways of going about this.

The secret in combinatorics is to look for the short cuts to the answer. There are usually different equivalent ways of enumerating things. In this case, the quick way to the answer is to count up the options you're not allowed to have.

Think of it this way. Every possible combination of digits is allowed (each of seven digits with 5 choices) except for a certain set. If you can figure what is the set that isn't allowed, you can just subtract it from the total.

I won't give away the next step but ask yourself what combinations aren't allowed and then work out how many of them there are.

- #7

Mithal

- 28

- 0

I hope this is the correct answer to your question :

(5 ) (5) * 5 ! - 5 =

(25) ( 120) - 5 = 2995 ways

(5 ) (5) * 5 ! - 5 =

(25) ( 120) - 5 = 2995 ways

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- #8

Jurij

- 14

- 1

Why is it correct answer? Could you explain it? I think it is too small.

- #9

neurocomp2003

- 1,366

- 3

Yes it is P&C problem;

well first off given only 5 distinct digits(i assume that's what you meant and let's call it alphabet to distinguish from 7digits)

[0]how many 7 digit numbers can you make with 5 digits(- - - - - - -) =

that is the total you can have. REMEMBER TO GET RID OF REPETITIONS

find an easy way to do this..hint if (a is the first digit then a must also be the last.

[1][Now what is the exclusion(well you want at least 2 different alphabets): you don't want any 7digit that has exactly only 1 alphabet present...

thus your exclusion set is #of 7digits of only 1 alphabet present...THUS YOU GOTTA FIGURE HOW MANY THERE ARE

Your solution will be #=TOTAL-Exlusion set.

lastly becareful how many times your exclusion set appears in the set of repetitions.

well first off given only 5 distinct digits(i assume that's what you meant and let's call it alphabet to distinguish from 7digits)

[0]how many 7 digit numbers can you make with 5 digits(- - - - - - -) =

that is the total you can have. REMEMBER TO GET RID OF REPETITIONS

find an easy way to do this..hint if (a is the first digit then a must also be the last.

[1][Now what is the exclusion(well you want at least 2 different alphabets): you don't want any 7digit that has exactly only 1 alphabet present...

thus your exclusion set is #of 7digits of only 1 alphabet present...THUS YOU GOTTA FIGURE HOW MANY THERE ARE

Your solution will be #=TOTAL-Exlusion set.

lastly becareful how many times your exclusion set appears in the set of repetitions.

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- #10

Mithal

- 28

- 0

5^7

excluding five choices (a,a,a,a,a) or (b,b,b,b,b) ...five of them

then the answer = 5^7-5=78120 ways

I am sorry again I made mistake.I hope this time it is right

- #11

Brinx

- 70

- 1

If you mean you have to make a 5-number sequence when you have 7 different numbers to choose from for each position, you need to do (number of possibilities per position)^(number of positions). Subtract the illegal choices from that number, and you're done.

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- #12

neurocomp2003

- 1,366

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- #13

Mithal

- 28

- 0

neurocomp2003 said:

I think it is right because these five choices are the only non - permissable in the condition of the problem ( at least two allowed different digits ) suppose else (a,a,a,a,b) is included a,b are two different digits .

- #14

neurocomp2003

- 1,366

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- #15

Mithal

- 28

- 0

neurocomp2003 said:

It is 5 ^7 .Every time you have to make five choices for each digit of the seven digits .repetion is allowed .So in this way you cover all possibilities and then subtract the five cases that not allowed in the condition of the problem as I mentioned before .

- #16

neurocomp2003

- 1,366

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for examples in 3digit words 121 and 121 are a repetition but in 5^3 they are counted separatedly so 2x rather than 1x.

- #17

Mithal

- 28

- 0

I didn't mean repeation in the sense you are talking about . I mean if you choose 3 for the first digit then you can use 3 again for the second digit or the third or etc. or any other number . That mean once you choose that number for that digit you can repeat it again subsequently for other digits.You can visualize this if you draw a tree and consider different cases .neurocomp2003 said:

for examples in 3digit words 121 and 121 are a repetition but in 5^3 they are counted separatedly so 2x rather than 1x.

I also would like to point out to a small mistake I made with the notations

I assumed the five digits are (a , b , c ,d ,e ) and

I meant ( a , a ,a ,a ,a ,a ,a ) and not like in the solution

( a , a ,a ,a ,a ) since it is seven digit number not five and the same for the others . But this has no effects on the logic of the answer . All stay the same .

- #18

neurocomp2003

- 1,366

- 3

i understood what repetition you were talking about. However i was not refereing to the alphabet(5digits) but rather the word(7digit word)

THe repetition i was thinking of was an undirectional ordering question where

abc is the same cba. I realized i was wrong. and i apologize.

The question was referring to directional words.

THe repetition i was thinking of was an undirectional ordering question where

abc is the same cba. I realized i was wrong. and i apologize.

The question was referring to directional words.

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