Option A:
The number of outcomes in which at least three heads appear in $6$ consecutive tosses of a fair coin.
A coin tossed $6$ times consecutively could result in outcomes ranging from $0$ heads to $6$ heads. We have to count the outcomes in which at least three heads appear;
What does at least three heads mean?
A minimum of $3$ heads. i.e., the outcomes in which we get $3$ heads or $4$ heads or $5$ heads or all $6$ as heads.
- Number of outcomes with $3$ heads: $6$ choose $3$ or $6 \mathrm{C} 3=6 ! / 3 ! \times 3 !=20$.
- Number of outcomes with $4$ heads: $6$ choose $4$ or $6 \mathrm{C} 4=6 ! / 4 ! \times 2 !=15$.
- Number of outcomes with $5$ heads: $6$ choose $5$ or $6 C 5=6 ! / 5 ! \times 1 !=6$.
- Number of outcomes with $6$ heads: $6$ choose $6$ or $6 C 6=1$ way.
So, the number of outcomes in which at least $3$ heads will appear $=20+15$ $+6+1=42$
Option A is NOT one of the answers.
Option B:
The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die is rolled thrice.
We can either have all three times Odd numbers, OR we can have exactly one Odd number.
- When we have all three times Odd numbers: $3 \times 3 \times 3=27$ Ways.
- When we have exactly one Odd number: $(3 \mathrm{C} 1) \times 3 \times 3 \times 3=81$
Final answer $=81+27=108$
Choice $B$ is one of the answers where the number of outcomes is more than $50.$
Option C:
The number of ways of posting $6$ different letters in $2$ different post boxes such that at least one letter is posted in each of the boxes.
Let us say the letters are addressed to $A, B, C, D, E$, and $F$, and the boxes are $1$ and $2 .$
Letter $A$ can be posted in the boxes in $2$ ways. Either into box $1$ or into box $2.$
Similarly, the letter $B$ and the remaining letters can be posted into a box in $2$ ways each.
Letter $A$ can be posted in $2$ ways AND letter $B$ can be posted in $2$ ways and so on.
So, the $6$ letters can be posted in $2 \times 2 \times 2 \times 2 \times 2 \times 2=26=64$ ways.
The only problem is that these $64$ ways include the possibility that all the $6$ letters are posted into only one of the post boxes - either all $6$ into box $1$ or all $6$ into box $2 .$
If we count these two possibilities out, then we have $64-2=62$ ways in which $6$ different letters can be posted into $2$ different post boxes such that at least one letter is posted in each of the boxes.
Option $\mathrm{C}$ is one of the choices in which the number of outcomes exceeds $50.$
Option D:
The number of outcomes in which the vowels appear together when the letters of the word 'PRIORITY' are reordered.
PRIORITY is an $8$-letter word.
- The vowels in the word are $\mathrm{I}, \mathrm{O}$, and $\mathrm{I}$.
- The consonants in the word are $\mathrm{P}, \mathrm{R}, \mathrm{R}$, $\mathrm{T}$, and $\mathrm{Y}$.
- The condition to be satisfied is that the vowels appear together.
So, if we consider the $3$ vowels as one unit, the vowels will appear together. Let us call this unit comprising $3$ vowels as $\delta$.
So, we need to reorder $\mathrm{P}, \mathrm{R}, \mathrm{R}, \mathrm{T}, \mathrm{Y}$ and $\delta$
i.e., $6$ letters have to be reordered - out of which two letters are 'R's.
So, these can be reordered in $6 ! / 2 !=360$ ways.
Now the unit $\delta$ comprising $3$ letters I, $O$, and I can also be reordered.
These $3$ letters of which two are alike, can be reordered in $3 ! / 2 !=3$
ways.
So, the total number of ways to reorder the letters of the word
PRIORITY such that the vowels appear together $=360 * 3=1080$ ways.
Option D is one of the choices in which the number of outcomes exceeds $50.$
