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Thurston wrote an important seven-digit [#permalink]
1
jelal123 wrote:
somebody pls explain .
regards


There are 7 numbers of which last 3 must have a 0, a number between 1-9 and a number between 0-9.
We can make 3 different cases:

Case I: 0 _ _
possible cases = 9 x 10 = 90

Case II: _ 0 _
possible cases = 9 x 10 = 90

Case III: _ _ 0
possible cases = 9 x 10 = 90

Total number of possible cases = 90 + 90 + 90 = 270

Since, 10 numbers were dialed from 270 cases, required probability = 10 /270 = 1/27

Hence, option C
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Re: Thurston wrote an important seven-digit [#permalink]
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POSSIBLE COMBINATIONS FOR LAST THREE DIGITS IS
1.NNN = 9*9*9 = 729
2.ZNN
3.ZZN
4.ZZZ=1*1*1 = 1
ACC TO CONDITIONS GIVEN 1 AND 4 IS NOT POSSIBLE..SO SUBTRACT THOSE FROM TOTAL POSSIBLE ARRANGEMENT OF 3 DIGITS
1000-729-1 = 270
SO 270 NUMBERS SATISFY OUR CONDITIONS NOW PERSON CHOOSES 10 OF THIS..PROBABILITY= 10/270=1/27
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Re: Thurston wrote an important seven-digit [#permalink]
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The cases can be NN0 or N00
= 9C2*3! (2 numbers chosen out of 9 and these three can be arrannges in 3 positions) + 9C1 * 3! (1 numbers chosen out of 9 and these three can be arrannges in 3 positions)
=216 + 54
=270
probability = 10/270 = 1/27
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Re: Thurston wrote an important seven-digit [#permalink]
Hello from the GRE Prep Club BumpBot!

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