# permutation

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• Jun 24th 2010, 05:37 AM
nikk
permutation
hai...i need help for the following basic problem. hope some on can guide me...

Question:

Digital devises store information in the form of digits using only the digits 0 or 1. Calculate the number of codes which can be stored if they consist of 10 digits.

Ans:
1024

• Jun 24th 2010, 05:46 AM
Ackbeet
Is that your answer? If so, how did you arrive at it?
• Jun 24th 2010, 05:49 AM
nikk
Quote:

Originally Posted by Ackbeet
Is that your answer? If so, how did you arrive at it?

no..that the answers from book. i can't get it. i need the help in order to get the ans.

• Jun 24th 2010, 05:55 AM
Ackbeet
Ok. So what thoughts have you had?
• Jun 24th 2010, 06:04 AM
nikk
Quote:

Originally Posted by Ackbeet
Ok. So what thoughts have you had?

what should i list..

is it 0 1 0 1 0 1 0 1 0 1 or 0 1 2 3 4 5 6 7 8 9....??

that is my big problem...

huwahuwa
• Jun 24th 2010, 06:06 AM
Ackbeet
You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?
• Jun 24th 2010, 06:11 AM
nikk
Quote:

Originally Posted by Ackbeet
You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?

ok2..let me try..hope u still there

heheeh
• Jun 24th 2010, 06:19 AM
nikk
Quote:

Originally Posted by Ackbeet
You have ten digits, each of which could be 0 or 1. So one possibility would be 0000000000. Another would be 0000000001. Then you could have 0000000010. And so on. How many ways could you have these bits flipped?

heheeh..i still stuck...i can't list it actually it will end until 1024 heehhe ( i refer to the answer) . so how to apply permutation here. what i can see it is a permutation which is involve semilar object..so nPn or nPr can't be use here...

hope can u gauide me
• Jun 24th 2010, 06:24 AM
Ackbeet
Let's suppose you only had two bits. How many ways can you set the first bit? How many ways can you set the second bit? So, how many ways can you set both bits?
• Jun 24th 2010, 06:41 AM
nikk
Quote:

Originally Posted by Ackbeet
Let's suppose you only had two bits. How many ways can you set the first bit? How many ways can you set the second bit? So, how many ways can you set both bits?

i had try but the ans is exceed the given ans in the book.
• Jun 24th 2010, 06:45 AM
Ackbeet
I'm not interested in the book's answer. I'm interested in your answer. You need to think, here! How many ways can you set two bits?
• Jun 24th 2010, 06:57 AM
nikk
Quote:

Originally Posted by Ackbeet
I'm not interested in the book's answer. I'm interested in your answer. You need to think, here! How many ways can you set two bits?

CASE 1
---------------------------------------------------------------------
for 0000000001 , the number of way is=10!/(9!x1!)=10
for 0000000010, the number of way is=10!/(9!x1!)=10
.
.
for 1000000000, the number of way is=10!/(9!x1!)=10

therefor the total the number of way is 10 x 10 = 100
--------------------------------------------------------------------
CASE 2

for 0000000011 , the number of way is=10!/(8!x2!)=45
therefore the total for case 2 is 45 x10 = 450
------------------------------------------------------------------
CASE 3
for 0000000111 , the number of way is=10!/(7!x3!)=120
therefore the total for case 3 is 120 x10 = 1200

if i add Case 1 to case 3 the ans already > ans in the text book

huwahuwa...
• Jun 24th 2010, 07:03 AM
Ackbeet
I'm afraid you're not thinking about the problem correctly. For each digit, you don't have 10 possibilities, you only have 2. Also, the possibilities don't add, they multiply. Just think about the two-bit case: you've got 00, 01, 10, 11. That's it. 4 possibilities. For 3 bits, you've got 000, 001, 010, 011, 100, 101, 110, 111. That's 8 possibilities. For 4 bits, you have 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111; that's 16 possibilities. Do you see how this is working?
• Jun 24th 2010, 07:13 AM
nikk
Quote:

Originally Posted by Ackbeet
I'm afraid you're not thinking about the problem correctly. For each digit, you don't have 10 possibilities, you only have 2. Also, the possibilities don't add, they multiply. Just think about the two-bit case: you've got 00, 01, 10, 11. That's it. 4 possibilities. For 3 bits, you've got 000, 001, 010, 011, 100, 101, 110, 111. That's 8 possibilities. For 4 bits, you have 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111; that's 16 possibilities. Do you see how this is working?

tq for ur reply, totally im wrong..i will try it now
• Jun 24th 2010, 07:44 AM
nikk
update:
i have try until bit no 5

2 bit, they have 4 ways
3 bit, they have 8 ways
4 bit, they have 16 ways
5 bit, they have 28 ways.

i just stop and do the multipication until 5 bit to check the ans as below
4x8x16x28>ans in the text book

so how??
any idea.. is it we need to add?
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