UPES B.Tech Admissions 2025
ApplyRanked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements
50 Questions around this concept.
Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs that can be formed such that is empty, is
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to
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 5C1 + 5C2 + … + 5C5 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.
"Stay in the loop. Receive exam news, study resources, and expert advice!"