Question

The number of ways of selecting three numbers out of first 50 natural numbers such that the
minimum difference of any two of the selected numbers is 5, is​

Answer 1

The number of ways of selecting three numbers out of first 50 natural numbers such that the minimum difference of any 2 of the selected numbers is 5, is 120.

• Given,
• 3 numbers out of first 50 natural numbers
• but the minimum difference of any two of the selected number is 5, is
• ⇒
• $\dfrac{50}{5}=10$
• only 10 numbers remained.
• Now,
• the number of ways of selecting 3 numbers
• $=^{10}C_3\\\\=\dfrac{10!}{(10-3)!*3!}\\\\=120$

Answer 2

The number of possible ways are 120

Explanation:

It is given that the number of ways of selecting 3 numbers from the first 50 natural numbers such that the  minimum difference of any two of the selected numbers is 5.

Thus, we have,

$\frac{50}{5}=10$

To select the 3 numbers from these 10 numbers, hence, we have,

$10 C_{3}$

Simplifying, we get, the result,

$\frac{10 !}{(10-3) ! \times 3 !}$

$\frac{10 !}{7 ! \times 3 !}$

$\frac{10\times9\times8\times7!}{7 ! \times 3 !}$

Cancelling the common terms, we get,

$10 C_{3}=120$

Hence, The number of possible ways are 120

Learn more:

(1) brainly.in/question/8959040

(2) brainly.in/question/6695330