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GRE Quantitative Reasoning - Advanced Concepts & Strategies

Advanced flashcard deck covering hard-level GRE quantitative topics including complex functions, multi-variable statistics, geometric properties, and strategic reasoning.

20 cards

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#1

Front

Define the number of non-negative integer solutions for the equation $x_1 + x_2 + \dots + x_k = n$.

Back

The number is given by the combination formula $C(n+k-1, k-1)$. This is derived using the 'Stars and Bars' theorem, where $n$ indistinct items (stars) are placed into $k$ distinct bins (dividers). For example, the equation $a + b + c = 10$ has $C(10+3-1, 3-1) = C(12, 2) = 66$ non-negative integer solutions.

#2

Front

What is the geometric interpretation of the determinant of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$?

Back

The absolute value of the determinant $|ad - bc|$ represents the area of the parallelogram formed by the vectors $(a, c)$ and $(b, d)$ in the Cartesian plane. If the determinant is zero, the vectors are linearly dependent (collinear), and the area collapses to zero.

#3

Front

Sum of the first $n$ terms of a geometric series ($S_n$) vs. Sum to infinity ($S_{\infty}$).

Back

$S_n = \frac{a_1(1-r^n)}{1-r}$. The sum converges to a finite limit only if the common ratio $r$ satisfies $-1 < r < 1$. In that case, as $n \to \infty$, $r^n \to 0$, so $S_{\infty} = \frac{a_1}{1-r}$. If $|r| \ge 1$, the series diverges (sum is infinite or undefined).

#4

Front

Effect of Outliers on Mean vs. Median.

Back

The arithmetic mean is highly sensitive to outliers because it incorporates every value in the dataset. The median is robust against outliers because it depends only on the middle value(s) of the ordered dataset. In a skewed distribution (e.g., income), the mean is typically pulled in the direction of the tail (skew) relative to the median.

#5

Front

Relationship between Permutations and Combinations regarding order.

Back

Permutations ($P(n, k)$) count arrangements where order matters: $\frac{n!}{(n-k)!}$. Combinations ($C(n, k)$) count selections where order does not matter: $\frac{n!}{k!(n-k)!}$. The relationship is $P(n, k) = C(n, k) \times k!$. For every combination of $k$ items, there are $k!$ ways to permute them.

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