If \({ }^{189} \mathrm{C}_{35}+{ }^{189} \mathrm{C}_{\mathrm{x}}={ }^{190} \mathrm{C}_{\mathrm{x}}\) then \(\mathrm{x}\) is equal to (a) 34 (b) 35 (c) 36 (d) 37

Recall that the combination formula is given by: \({ }^nC_r = \frac{n!}{r!(n-r)!}\), where n is the number of elements, r is the number of elements we want to select and ! denotes a factorial. Now, let's express the given equation in terms of factorials. \({ }^{189}C_{35} + { }^{189}C_{x} = { }^{190}C_{x}\) \(\frac{189!}{35!(189-35)!} + \frac{189!}{x!(189-x)!} = \frac{190!}{x!(190-x)!}\)

Now, let's simplify the equation further by cancelling out 189! in both the left-hand side and right-hand side terms: \(\frac{1}{35!(154)!} + \frac{1}{x!(189-x)!} = \frac{190}{x!(190-x)!}\)

Considering that both sides of the equation are equal, we can set the numerators and denominators equal and compare them: Denominators comparison: \(35!(154)! = x!(189-x)!\) Numerators comparison: \(190 = 1 + 1\)

As we continue with working with our denominators comparison, we can divide both sides by 154!. \(35! = x!(35-x)!\) Taking into account that there is only one value on the right side of the numerators comparison, we see that \(1+1=2\), let's now use that in denominators from both sides of the equation: \(2 \times 35! = 190 \times x!(35-x)!\) Divide both sides by 35! to isolate x: \(2 = 190 \times \frac{x!(35-x)!}{35!}\) Now, let's compare the answer choices with the simplified equation: (a) 34: \(2 = 190 \times \frac{34!(1)!}{35!}\) \(2 = 190/35\), this is not true. (b) 35: \(2 = 190 \times \frac{35!(0)!}{35!}\) \(2 = 190\), this is not true. (c) 36: \(2 = 190 \times \frac{36!(35-36)!}{35!}\) \(2 = 190 \times \frac{36}{35}\) \(70 = 190\), this is not true. (d) 37: \(2 = 190 \times \frac{37!(35-37)!}{35!}\) \(2 = 190 \times \frac{1}{35}\) \(70 = 190\), this is true. Therefore, the correct answer is (d) 37.