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Let there be m of 25p coins, and p number of 5paise.

Then there are n = 36 - m - p number of 10p coins.

Assume that there is at least 1 of each type of coin.

Their total value in paise is :

25 m + 10 n + 5 p = 300 paise

25 m + 10 (36 - m - p) + 5 p = 300

5 p = 15 m + 60

p = 3 m + 12

As m >= 1, p >= 15 and p is a multiple of 3. m , n , p are all integers.

m = 1 p = 15 n = 36 - 15 - 1 = 20

m = 2 p = 18 n = 16

m = 3 p = 21 n = 12

m = 4 p = 24 n = 8

m = 5 p = 27 n = 4

So there are five combinations of coins possible.

Then there are n = 36 - m - p number of 10p coins.

Assume that there is at least 1 of each type of coin.

Their total value in paise is :

25 m + 10 n + 5 p = 300 paise

25 m + 10 (36 - m - p) + 5 p = 300

5 p = 15 m + 60

p = 3 m + 12

As m >= 1, p >= 15 and p is a multiple of 3. m , n , p are all integers.

m = 1 p = 15 n = 36 - 15 - 1 = 20

m = 2 p = 18 n = 16

m = 3 p = 21 n = 12

m = 4 p = 24 n = 8

m = 5 p = 27 n = 4

So there are five combinations of coins possible.