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The question is - Out of 2 grapes, 3 mangoes & 4 apples, how many different selections can be made if each selection has at least one fruit of every kind?

Now, they have solved the problem using the following method

No. of ways = $(1+1)(1+1+1)(1+1+1+1)$ i.e. (one way of selecting 1 grape + one way of selecting 2 grapes)(...)(...)

1 because they are identical

My question is- Why can't I use the following?

$(1+1+1)(1+1+1+1)(1+1+1+1+1)$-1 i.e(1 way of selecting no grapes+ same as above)(...)(...)-1(i.e 1 way when no fruits where chosen)

If I do this, the answer is not matching?

Kindly assist me!

2 Answers 2

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Your approach counts the ways in which selections can be made if at least one fruit is to be selected.

The first part, $(1+1+1)(1+1+1+1)(1+1+1+1+1)$ counts number of ways of selection without any restrictions. This selection can contain $0$ fruit, $1$ fruit, ... all the $9$ fruits.

Then you deduct $1$ to avoid selection of $0$ fruits.

When you calculate $ (1+1)(1+1+1)(1+1+1+1)$ , we are making sure at least one from each kind of fruit is included and this is what needed in the question.

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    I see how it works! – 2017-02-23
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If the apples (similarly mangoes, grapes) are considered identical, in the sense that a selection with apples A and B is not considered different than a selection with apples C and D, then the answer $2\cdot 3\cdot 4$ is indeed right: You can select $1$ or $2$ grapes, $1, 2$ or $3$ manoges, and $1, 2, 3$ or $4$ apples.

On the other hand,m if they are distinguishable, then there are for example $2^4-1 = 15$ ways to choose at least one apple, and the answer will be $(2^2-1)(2^3-1)(2^4-1)$ different selections.