Question

There are 4 mangoes, 3 apples, 2 oranges and 1 each of 3 other varieties of fruits. The number of ways of selecting at least one fruit of each kind is.

Answer 1

Hint:
In Mathematics whenever we need to choose r items out of n items we use combinations. This is represented by ${}^{n}{C_r}$ and it’s formula is $\dfrac{n!}{r!(n-r)!}$. we’ll use the same concept to choose and count the ways.

Complete step by step solution:
To find the number of ways of selecting at least one fruit of each kind
Number of ways of selecting $$1$$ mango from $$4$$ mangoes is $${}^4{C_1}=\dfrac{4!}{1!(4-1)!} \Rightarrow 4$$
Number of ways of selecting $$1$$ mango from $$3$$ apples is $${}^3{C_1}=\dfrac{3!}{1!(3-1)!} \Rightarrow 3$$
Number of ways of selecting $$1$$ mango from $$2$$ oranges is $${}^2{C_1}=\dfrac{2!}{1!(2-1)!} \Rightarrow 1$$
There is only one way selecting one fruit from one.

$$\therefore $$ Total no of ways of selection $$ = 4 \times 3 \times 2 \times 1 \times 1 \times 1 = 4!$$

Note:
Students many times get confused between permutations and combinations. Keep it in mind, when we need to count the number of ways to rearrange the items then we’ll use permutations and combinations when we need to count the number of ways of choosing some items.
Formula for $n!$ is $n \cdot (n-1)!$