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How many three digit numbers are there whose first digit equals the sum of the second and third digits?
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Let the number be xyz

1) $x$ is 9, so $y+z=9$ Ways : $(9,0) ;(8,1) ;(7,2) ;(6,3) ;(5,4)$..
Each of the above 5 ways can be taken as 2 different arrangements.. 981 or 918 So $5^{*} 2=10$

2) $x$ is 8 , so $y+z=8$ Ways : $(8,0) ;(7,1) ;(6,2) ;(5,3) ;(4,4)$..
4 of the above 5 ways can be taken as 2 different arrangements..
So $4^{*} 2+1=9$

3) $x$ is 7 , so $y+z=7$ Ways : $(7,0) ;(6,1) ;(5,2) ;(4,3)$
Each of the above 4 ways can be taken as 2 different arrangements.. So $4^{*} 2$

4) $x$ is 6, so $y+z=6$ Ways : $(6,0) ;(5,1) ;(4,2) ;(3,3) \ldots$
3 of the above 4 ways can be taken as 2 different arrangements..
So $3^{*} 2+1=7$
And so on There is a pattern... So $10+9+8+7+6+5+4+3+2=54$
Last 2 is when $x$ is $1 \ldots$ 101 or 110
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