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How many 3-digit numbers can be formed so that the units place of a th

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New post Updated on: 30 Oct 2018, 11:05
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How many 3-digit numbers can be formed so that the units place of a three-digit number is always prime? Repetition of digits is not allowed.

Can someone explain this with the 'fill in the blanks with number of available options' method?





OA: 256

Originally posted by manavivarma on 30 Oct 2018, 09:30.
Last edited by manavivarma on 30 Oct 2018, 11:05, edited 1 time in total.
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New post 30 Oct 2018, 10:00
manavivarma wrote:
How many 3-digit numbers can be formed so that the units place of a three-digit number is always prime? Repetition of digits is not allowed.

Can someone explain this with the 'fill in the blanks with number of available options' method?


manavivarma

Can you update the options and OA in the question
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New post 30 Oct 2018, 11:04
Princ there were no options to this question. Have updated the OA in my post though :)
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New post 30 Oct 2018, 11:07
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8 * 8 * 4 = 256 ways
Prime no between 0-9 : 2 , 3 , 5 , 7
Total digits between 0-9 = 10
So unit digit can be filled in 4 ways
Now 9 digits remain. First slot cannot start with 0 otherwise it won't be a 3 digit number
So first slot can be filled in 8 ways.
Middle slot can have 0 so it can be filled in 8 ways.
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New post 30 Oct 2018, 11:09
Thanks pandeyashwin!
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New post 30 Oct 2018, 11:22
Three Digit numbers from total 10 digit (0-10)
Restriction: Unit digit has to be prime. i.e. Either 2,3,5,or 7

Suppose three digit is A-B-C
Case 1: Unit digit 2:
We have 1 option for C (i.e.-2)=1
We have 8 option for A ( 1,3,4,5,6,7,8,9)=8
We have another 8 option for B (except 2 and digit taken by A)=8
So total 8*8*1=64 ( for unit digit 2)

we have total 4 cases.


64*4=256

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New post 30 Oct 2018, 11:23
Thanks for updating OA

For units place, We have \(4\) choices i.e \(2,3,5,7\)
For hundreds place, We have \(8\) choices \([10(0 \quad to \quad 9)-1\)(Hundreds place cannot be \(0\))-\(1\)(Digit already used for units place)]
For Tens place, We have \(8\) choices \([10(0 \quad to \quad 9)-1\)(Digit already used for Hundreds place)-\(1\)(Digit already used for units place)]
Total Number of 3 digits such that units place of a the number is prime number and having distinct digits:\(8*8*4=256\)
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Re: How many 3-digit numbers can be formed so that the units place of a th  [#permalink]

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New post 10 Apr 2019, 11:09
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But what if after filling up the units place , we fill up the tens place. And only then the hundreds place .
Total 10 digits { 0 to 9 }

In that case wouldn't it be
4(units place = 2,3,5,7) * 9 (tens place = {0to9} minus 1 digit which has been placed at units place ) * 7(hundreds place = all digits except 0 , units place digit and tens place digit
Therefore , 4*9*7 ?

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Re: How many 3-digit numbers can be formed so that the units place of a th   [#permalink] 10 Apr 2019, 11:09
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