Problems on Counting

Question: The number of surjective maps from a set of $4$ elements to a set of $3$ elements is

$1.~36 \quad 2.~64 \quad 3.~69 \quad 4.~81$

Answer: Let $A$ be a $4$ elements set and $B$ be a $3$ elements set. Here we need to find number of surjective maps between $A$ onto $B$. The idea is to partition the set $A$ into $3$ non-empty subset and assign each subset to a element of $B$.
Different partition of set $A$ into $3$ non-empty subset is equal to $\binom{4}{2}$, it can be represented by following figure(more generally it is given by Starling number of second kind)

Thus we got a total of $6$ partition. Each partition corresponds to $3!$ surjective maps. Hence total number of surjective maps $= 6 \times 3! = 36$.

Hence 1 is correct choice.

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Question: An ice cream shop sells ice cream in five possible falvours: Vanilla, Chocolate, Strawberry, Mango and Pineapple. How many combinations of three scoop cones are possible? [Note: The repeatation of flavours allowed but the order in which the flavours are chosen does not matter.]

$1.~10 \quad 2.~20 \quad 3.~35 \quad 4.~243$

Answer: Let’s count the possibilities one by one
1. All 3 cones are of same flavor, we have $5$ possiblities.
2. All 3 cones are of different flavor, we have $\binom{5}{3} = \frac{5!}{3!\cdot2!} = 10$ possibilities.
3. 2 are of same flavor and 1 of different flavor, then we have $5 \times 4 = 20$ possibilities.
Hence total choice is $5+10+20 = 35$.

Hence 3 is correct choice.

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Question: We are given a class consisting of $4$ boys and $4$ girls. A committee that consists of a President, a Vice-President and a Secretary is to be chosen among the $8$ students of the class. Let $a$ denote the number of ways of choosing the committee in such a way that the committee has at least one boy and at least one girl. Let $b$ deonote the number of ways of choosing the committee in such a way that the number of girls is greater than or equal to that of the boys. Then

$$\begin{align*}&1. ~a = 288 &2. ~a = 168 \\ &3.~b = 144 &4.~b = 192 \end{align*}$$

Answer: First we will choose a group of three people satisfying the condition and then multiply by $3!$, since the places are distinguishable.

Committee a: Choosing at least one boy and one girl = (Choosing 2 boy and 1 girl) + (Choosing 1 boy and 2 girl) = $2\binom{4}{2}\binom{4}{1}= 48$.
Hence no of ways choosing comittee a = $48 \times 6 = 288$.

Committee b: Choosing a committee in such a way that the number of girls is greater than or equal to that of the boys = (Choosing 2 girl and 1 boy)= $\binom{4}{2}\binom{4}{1}= 24$.
Hence no of ways choosing comittee a = $24 \times 6 = 144$.

Hence 1 and 3 are correct options.