Barisal HS 2009_12

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rakeen
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Barisal HS 2009_12

Unread post by rakeen » Mon Jul 11, 2011 12:40 pm

A student has a five digited id card number. If you choose a student, what is the probability of 'not' having same number on it(i.e. 22056, there's 2)?

There are 100000 such id cards. Any digit for example '0' can be placed in the number in $2^5$ ways. But of them, there are 5+1=6 such ways where '0' is only one. So, it is not acceptable. So, $2^5 -6 = 26$ . For each digit(0-9) it is 26*10 = 260. So the probability is, (260/100000)*100% = .26% Am I right?{it seems not :( }
r@k€€/|/

Mehfuj Zahir
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Re: Barisal HS 2009_12

Unread post by Mehfuj Zahir » Mon Jul 11, 2011 7:18 pm

just use the concept of box. let you 5 boxes and you want to fill the box with different digits
now you have 9 ways to fill the 1st box as the id card number is 5 digit number you dont use 0 on it and now you have also 9 ways to fill the 2nd box and 8 ways for 3 rd boxs so the result is 9.9.8.7.6. and you have 99999 five digit number ( i use the id numbers are five digit number if there exist this type of id number 01973 there will be slight change in result)

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Labib
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Re: Barisal HS 2009_12

Unread post by Labib » Fri Dec 02, 2011 9:44 pm

শিশির ভাই, আইডি কার্ড এ তো প্রথম অঙ্ক ০ হতেও পারে??
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nafistiham
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Re: Barisal HS 2009_12

Unread post by nafistiham » Sat Dec 03, 2011 6:04 pm

yes, i also think the same way.if it is, the answer will just be
\[\frac{(10-5)!}{10^6}\]
of course, Mehfuj Zahir has mentioned about it,too.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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bristy1588
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Re: Barisal HS 2009_12

Unread post by bristy1588 » Mon Dec 05, 2011 5:34 pm

Proshno ta bujhi nai...Labib Rashid.. Bujhai dao
Bristy Sikder

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