The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to ______.
The number of four-digit numbers that can be formed using the digits 2, 1, 2, and 3 is \( \frac{4!}{2!} = 12 \).
These are the permutations of the digits 2, 1, 2, and 3. The sum of digits at the unit place is calculated as: \[ 3 \times 1 + 6 \times 2 + 3 \times 3 = 24. \] Now, the required sum is: \[ 24 \times 1000 + 24 \times 100 + 24 \times 10 + 24 \times 1 = 24 \times (1000 + 100 + 10 + 1) = 24 \times 1111 = 26664. \] Thus, the sum is \( 26664 \).
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:
\(^nP_r = \frac{n!}{(n-r)!}\)
nPr = permutation
n = total number of objects
r = number of objects selected