Question

The number of $3$ digit numbers of the form $xyz$ where $x,z<y$ and $x\ne 0$ is

• A. $250$
• B. $240$
• C. $260$
• D. $280$

Answer 1

The correct option is B $240$

We have to find the number of $3$ digit numbers of the form $xyz$ where$x,z<y:$
Let $0$ be allowed in the $1^{st}$ position.

Let all digits be distinct.
Then, number of ways of selecting $3$ digits $={}^{10}{C}_3=120$
Of the $3$ digits (all distinct), the maximum takes the middle position while the other $2$ digits take the $1^{st}$ and $3^{rd}$ positions.

However, these $2$ positions may be interchanged.
This can be done in $2!=2$ ways
So, total ways = $120\times 2=240$

Now, we need to remove all numbers which have started with $0$ and have $y>z$, since a $3$ digit number cannot start with $0$.

For this, we need to select $2$ digits from $1-9$ (since we have taken $x=0$, so $z$ cannot be $0$, since $x,y,z$ are distinct as assumed above) and for each selection, there can only be $1$ way of forming the number.

Thus, number of ways $={}^9{C}_2=36$.

Hence, number of ways $=240-36=204$.

Now, let $x$ and $z$ are identical and $y$ is different.

Thus, none among $x,y$ nor $z$ can be $0$. So, we need to select $2$ numbers from $1-9$ and for each selection, we can only have one number.

This can be done in ${}^9{C}_2=36$ ways.

Hence, the answer $=204+36=240$

Hence, (b) is correct.