By using all numbers, i.e. 123456789 and subtraction/addition, operators number 100 can be formed in many ways.
Example 98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100

But if we add a condition that use of the number 32 is the must. Then there are limited solutions.
One of such solution is: 9 - 8 + 76 + 54 - 32 + 1 = 100

Can you tell any other solution?

9 - 8 + 7 + 65 - 4 + 32 - 1 = 100
9 - 8 + 76 + 54 - 32 + 1 = 100

Other ways of making number 100 (withiout using number 32)
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100

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