Question

How many three-digit positive integers, with digits x, y, and z in the hundred’s, ten’s and unit’s place respectively exist such that \[x< y, z (a) 245
(b) 285
(c) 240
(d) 320

Answer 1

Hint: We are asked to make 3 digit positive integer with x, y and z. We will first look at which one is greatest and which one is smaller. We have x < y and z < y which gives us that y is the greatest of all. Then we will start putting value for y and look to find the possible values of x and z. From these possible values, we will get the possible 3 digit number that can be formed. We will add all the possible 3 digit numbers to get our solution.

Complete step-by-step solution:
We are given that we have to make a 3 digit number with x at the hundredth place, y at the tenth place and z at one’s place. Each x, y and z are integers, so it can take value from 0 to 9. We have certain conditions for x, y and z. Firstly, we have x < y and then z < y which gives us that y is the greatest of all. Secondly, \[x\ne 0,\] means that x cannot be 0. So, we have that x and z are dependent on the value of y and also x cannot take 0.
Now, we will look at the various cases which arise due to different value of y.
(i) Now when the value of y = 0. As x < y, so the only possibility for x is negative numbers. But we need positive 3 digit numbers. So, no value possible for x and hence no number is formed when y = 0.
(ii) When value of y = 1. Again x < y, as y = 1, so the only possibility for x is 0 or negative number. And x cannot take zero and we need positive terms. So, no value is possible for x when y = 1. Hence no number is formed.
(iii) When value of y = 2. Now, x < y and x > 0 (As x cannot take value 0). So, the possible value of x is 1. For z, we have z < y. So, z can take two values 0 and 1. This means for y = 2, we can have x = 1 and z = 0, 1. So, total possibilities = \[1\times 2.\]
(iv) When value of y= 3. As y = 3 and we know that x < y and x > 0. So, x can take the value 1 and 2. As z < y, so, for z, z can take 0, 1, 2. This means that for y = 3, we have x = 1, 2 and z = 0, 1, 2. So, total possibilities \[=2\times 3.\]
(v) When value of y = 4. We have x < y and \[x\ne 0,\] so x can take the value 1, 2 and 3. And we have, z < y. So, y can take the value 0, 1, 2 and 3. This means that for y = 4, we have x as 1, 2, 3 and z as 0, 1, 2, 3. So, the total possibilities \[=3\times 4.\]
In the same manner, we keep on going.
We get when y = 9, the value of x can be 1, 2, 3, 4, 5, 6, 7, 8 and for z values can be 0, 1, 2, 3, 4, 5, 6, 7, 8.
Therefore, the total possibilities will be \[8\times 9.\]
Now, to find the total number of 3 digit numbers formed, we will add up these possibilities.
Hence, a total of 3 digit numbers formed will be
\[=1\times 2+2\times 3+3\times 4+4\times 5+5\times 6+6\times 7+7\times 8+8\times 9\]
\[=2+6+12+20+30+42+56+72\]
Solving, we get,
Total 3 digit number formed = 240
Hence, the option (c) is the right answer.

Note: For y = 2, we get x can have value 1 only and z can have value 0 or 1. When we make numbers, the value of y is fixed as y = 2 but x and z can change from the possibility of numbers they have. That’s why total possibility becomes \[1\times 2\] as x have 1 to choose from which z has 2 numbers to choose from, similarly for others.