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-1 + 2 × 0! =...?

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Answer
    1
Explanation

In questions of BODMAS or PEDMAS there is no any discussion on factorial sign. The exports of BODMAS are still not discuss in detail on it.

If you are not used to with functions, you can think about the square root and the trigonometric functions (sine, cosine, and tangent similarly cosecant, secant, and cotangent are the reciprocals of the basic trigonometric ratios) are examples of functions.

In questions of BODMAS or PEDMAS where logarithms, trigonometric functions, fictorials, expressions involving e etc are all treated as functions. This means that we have to solve them before multiplication, division, addition, or subtraction. Therefore we have to solve factorial sign value (0!) before doing any solution. i.e just convert 0! into normal number then solve multiplication, division, addition, or subtraction.



First of all we should understand about 0!. Here sign ! is read as factorial sign and 0! = 1. Technically, the factorial is also called unary operator. Just puts functions at the top of the order of operations, just after brackets.

      -1 + 2 × 0!  Solve the fictorial sign (0! = 1)
                   Just convert 0! into normal number 

    = -1 + 2 × 1   Solve the multiplication
    = -1 + 2       Solve the subtraction 
    = 1            

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Note

In above question we are applying BODMAS rule
B for bracket,
O for Order/Of,
D for Division,
M for Multiplication,
A for Addition,
S for Subtraction

ORDER OF OPERATION ([{(-)}], ()^2 √ , ÷, ×, +, -)

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  • 3 ÷ (5 ÷ 7) = (3 ÷ 5) ÷ 7
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  • Solve 4 - 2 ÷ (1/2) + 3 =...?
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  • Solve 2 - 2 × 2 - 2 ÷ 2 =...?
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  • Solve 3 - 3 × 3 - 3 ÷ 3 =...?
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  • ______________________

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  • ______________________

  • Which step is wrong?
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