Digits are 1,2,2,2,3,3,5
If unit digit 5 , then total numbers \(=\frac{6!}{3!2!}=60\)
If unit digit 3 , then total numbers \(=\frac{6!}{3!} =120\)
If unit digit 1 , then total numbers \(=\frac{6!}{3!2!}=60\)
∴ total numbers =60+60+120=240
So , the correct answer is 240
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , 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 $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.