Question

A box contains \[6\] apples, \[5\] oranges and \[8\] mangoes.
I.In how many different ways can a fruit be selected from the box?
II.In how many different ways can an apple and an orange be selected?

Answer 1

Hint: We can solve this problem using the concept of Combinations. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

Complete step by step solution:
I.Given that:
The number of apples in a box \[ = 6\]
The number of oranges in a box \[ = 5\]
The number of mangoes in a box \[ = 8\]
Then, the total number of fruits in a box is given by \[19\]
Let us find the number of different ways of selecting a fruit from the box is given as follows
Since we have to select one fruit out of 19 fruits and is \[{ = ^{19}}{C_1}\]
On simplification We get \[^{19}{C_1} = 19\]
Therefore, the number of different ways of selecting a fruit from the box is \[ = 19\] .
So, the correct answer is “19”.

II.Now Let us find the number of different ways of selecting an apple and an orange from the box and is given by
 \[{ = ^6}{C_1}{ \times ^5}{C_1}\]
 \[ = 6 \times 5\]
 \[ = 30\]
Therefore, the number of different ways of selecting an apple and an orange from the box is \[ = 30\]
So, the correct answer is “30”.

Note: In the above problem we have used the formula \[n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] to simplify the terms of the form \[^n{C_r}\] The symbol \[!\] is called factorial and the continued product of the first n natural number is called factorial ‘n’. It is denoted by \[n!\] or \[n! = n(n - 1)(n - 2)...3 \times 2 \times 1\] and also we know that \[0! = 1! = 1\]