Is that your answer? If so, how did you arrive at it?
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
tq for your advice.
CASE 1
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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
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CASE 2
for 0000000011 , the number of way is=10!/(8!x2!)=45
therefore the total for case 2 is 45 x10 = 450
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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...
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?
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?