If there are 50 notes whose total value is 100 rupees but 2 rupee note should not be there in the count of those50 notes How many such notes can be? If there are 50 notes whose total value is 100 rupees but 2 rupee note should not be there in the count of those 50 notes.How many such notes can be ?

Notes available are
$1$ Rupee
$2$ Rupees ( but NOT to use as per question )
$5$ Rupees
$10$ Rupees
$20$ Rupees
$50$ Rupees
$100$ Rupees ( obviously, this won't be needed!) 
$500$ Rupees ( obviously, this won't be needed!) 
$1000$ Rupees ( obviously, this won't be needed!)

Putting it other way

let us assume $a$,$b$,$c$,$d$ &amp; $e$ as variables satisfying following condition :
 
1. $a+b+c+d+e = 50$
2. $a + 5b + 10c + 20d + 50e = 100$

Then what can be values of $a,b,c,d$ &amp; $e$ ? ( Given that they must be non-negative integers!) 

Thanks!
One way to think about this is that the average note has value 2 rupees and only one note is any lower.  So each large note has to be paired with enough 1 notes to bring the average down to 2, and the group can be viewed as a "note".  A 5 rupee note has to be paired with 3 1 rupee notes making a package that is worth 8 rupees and has 4 notes.  Similarly a 10 rupee note must be paired with 8 1 rupee notes to make a package worth 18.
Your new notes have value 8, 18, 38, 98 and you need to make 100 (the count will take care of itself).  You can't use 98 at all.  Working mod 8 can make it easy.  If you use two 38's, you need 24 more, which can only be done with 3 8's.  So this gives 2*20+3*5+45*1 (in the original notes).  If you use one 38, that is 6 mod 8, so to get to 4 mod 8 we need 3 18's and 1 8, giving 20+3*10+5+45*1.  Finally, if we use no 38's, we need 2 18's and 8 8's, which leads to 2*10+8*5+40*1.
$a = 5, b = 1, c = 2, d = 1, e = 1$ hence $a + 5b + 10c + 20d + 50e$ will be $5 + (5\times 1) + (10\times 2) + (20\times 1) + (50\times 1) = 5 + 5 + 20 + 20 + 50 = 100$.