To ensure each child gets at least 2 apples, we can start by giving 2 apples to each of the three children.
Total apples given:
$2 \times 3 = 6$
Remaining apples:
$21 - 6 = 15$
Now, we need to distribute these remaining 15 apples among the 3 children with no additional restrictions (each child can get zero or more apples).
This problem now becomes a "distribution of identical items into distinct groups" problem.
We can use the stars and bars method to calculate the number of ways to distribute 15 identical apples among 3 children.
The formula for distributing $n$ identical items into $r$ distinct groups is:
\(n + (r-1)_{C_{r-1}}\)
Here, $n = 15$ (remaining apples) and $r = 3$ (children), so:
\(15 + (3-1)_{C_{3-1}} = ^{17}C_{2}\)
Now, calculating $^{17}C_{2}$:
$^{17}C_{2} = \frac{17 \times 16}{2} = 136$
Thus, the number of ways to distribute the 21 apples such that each child receives at least 2 apples is $136$.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \(^{8}P_{3} - ^{10}C_{3}\) | (I) 6 |
(B) \(^{8}P_{5}\) | (II) 21 |
(C) \(^{n}P_{4} = 360,\) then find \(n\). | (III) 216 |
(D) \(^{n}C_{2} = 210,\) find \(n\). | (IV) 6720 |
Choose the correct answer from the options given below:
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:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: