Question

The total number of selection of at least one fruit which can be made from $3$ bananas, $4$ apples and $2$ oranges is:
(A) $515$
(B) $511$
(C) $512$
(D) none of these

Answer 1

Hint: Given in the question that selection must be made with at least one fruit among the given fruits. To find the probability we must find the probability of selecting 1 fruit, 2 fruits, 3 fruits and so on and add the probabilities to get the final probability.

Complete step-by-step answer:
Given, at least one fruit will be selected from $3$ bananas, $4$ apples and $2$ orange.
Total number of fruits $ = 3 + 4 + 2 = 9$
Hence, at least one fruit will be selected from the total $9$ number of fruits.
Therefore, Number of selection of at least one fruit = Number of selection of $1$ fruit + Number of selection of $2$ fruits + Number of selection of $3$ fruits +…………….+ Number of selection of $9$ fruits
$ = {}^9{C_1} + {}^9{C_2} + {}^9{C_3} + {}^9{C_4} + {}^9{C_5} + {}^9{C_6} + {}^9{C_7} + {}^9{C_8} + {}^9{C_9}$
We know that, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$ = \dfrac{{9!}}{{1!\left( {9 - 1} \right)!}} + \dfrac{{9!}}{{2!\left( {9 - 2} \right)!}} + \dfrac{{9!}}{{3!\left( {9 - 3} \right)!}} + \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}} + \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}} + \dfrac{{9!}}{{6!\left( {9 - 6} \right)!}} + \dfrac{{9!}}{{7!\left( {9 - 7} \right)!}} + \dfrac{{9!}}{{8!\left( {9 - 8} \right)!}} + \dfrac{{9!}}{{9!\left( {9 - 9} \right)!}}$
$ = \dfrac{{9!}}{{1!8!}} + \dfrac{{9!}}{{2!7!}} + \dfrac{{9!}}{{3!6!}} + \dfrac{{9!}}{{4!5!}} + \dfrac{{9!}}{{5!4!}} + \dfrac{{9!}}{{6!3!}} + \dfrac{{9!}}{{7!2!}} + \dfrac{{9!}}{{8!1!}} + \dfrac{{9!}}{{9!0!}}$
$ = \dfrac{{9 \times 8!}}{{1 \times 8!}} + \dfrac{{9 \times 8 \times 7!}}{{2 \times 1 \times 7!}} + \dfrac{{9 \times 8 \times 7 \times 6!}}{{3 \times 2 \times 1 \times 6!}} + \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{4 \times 3 \times 2 \times 1 \times 5!}} + \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5! \times 4 \times 3 \times 2 \times 1}} + \dfrac{{9 \times 8 \times 7 \times 6!}}{{6! \times 3 \times 2 \times 1}} + \dfrac{{9 \times 8 \times 7!}}{{7! \times 2 \times 1}} + \dfrac{{9 \times 8!}}{{8! \times 1}} + \dfrac{{9!}}{{9!}}$
$ = 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1$
$ = 511$
Thus, there are $511$ ways of selection of at least one fruit which can be made from $3$ bananas, $4$ apples and $2$ oranges.

Hence, option (B) is the correct answer.

Note: The combination is a selection of a part of a set of objects or selection of all objects when the order doesn’t matter. Therefore, the number of combinations of $n$ objects taken $r$ at a time is given by the formula: ${}^n{C_r} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)........\left( {n - r + 1} \right)}}{{r!}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where the symbol denotes the factorial which means that the product of all the integers less than or equal to $n$ but it should be greater than or equal to $1$.
For example,
$0! = 1$
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$