Selection Of Any Number Of Objects - Practice Questions & MCQ

In certain situations, one has the liberty of selecting any number of objects from n (say) given objects. In this case, one can select 0 objects or 1 object or 2 objects or 3 objects or so on.... or all n objects.

Further, if the n objects are all different objects then not just how many objects are to be selected but a further question of which objects are selected also assumes importance. Thus there are two cases viz. the n objects being distinct or being identical.

Selections of any number of objects out of n DISTINCT objects:

Total no. of selections [Including Empty Selection]
$
{ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2 \ldots+\ldots{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=2^{\mathrm{n}}
$

Total no. of Non Empty selection $=2^n-1$
$
{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2+{ }^{\mathrm{n}} \mathrm{C}_2 \ldots+\ldots{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=2^{\mathrm{n}}-1
$

Example: A buffet dinner consists of 5 different dishes. In how many different ways can one help oneself if he has to take at least one dish?

Solution: The person can help himself to 1 or 2 or 3 or 4 or 5 dishes. Further, when he takes 1 or 2 or 3 or 4 or 5 , he can also choose which of the dish he takes. Thus he can help himself in ${ }^5 \mathrm{C}_1+$ ${ }^5 C_2+\ldots+{ }^5 C_5$ i.e. $32-1=31$ ways.

Selections of Any number of objects out of n IDENTICAL objects:

Total no. of selections [including Empty Selection] = n+1

Total no. of Non Empty selections = n ways

These both cases can be justified as selecting 1 or 2 or 3...or...n objects can be done in 1 way each (as each object is identical), so total n ways and if we don’t select any then it adds one more way of selecting 0 objects, hence n+1 ways

Question: In how many different ways can a person make a purchase from a fruit seller who has 5 mangoes, 8 apples and 10 oranges left with him and if the person has to purchase at least 1 mango, at least 1 apple and at least 1 orange?

Solution: Since at least 1 of each type has to be purchased, the number of ways with each of the different fruits can be purchased is 5 ways, 8 ways and 10 ways respectively. Thus, the total number of ways in which the purchase can be made is 5 × 8 × 10 = 400 ways.