In a group of students, 189 play Soccer, 129 play Basketball, 116 play Volleyball. 70 play Soccer and Basketball, 58 play Soccer and Volleyball, 51 play Basketball and Volleyball. If 339 students play

The formula for the number of elements in the union of three sets A, B, and C is:

$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$

We are given:
$n(A \cup B \cup C) = 339$
$n(A) = 189$
$n(B) = 129$
$n(C) = 116$
$n(A \cap B) = 70$
$n(A \cap C) = 58$
$n(B \cap C) = 51$
$n(A \cap B \cap C) = x$

Substitute these values into the formula:

$$339 = 189 + 129 + 116 - 70 - 58 - 51 + x$$

First, sum the individual set sizes:
$189 + 129 + 116 = 434$

Next, sum the sizes of the pairwise intersections:
$70 + 58 + 51 = 179$

Now, substitute these sums back into the equation:

$$339 = 434 - 179 + x$$

$$339 = 255 + x$$

Subtract 255 from both sides:

$$x = 339 - 255$$ $$x = 84$$

The value of $x$ is 84.