Hi Satish,

Technically speaking infinity is not a value, so in that case, we need to find the real values.

Hope this helps.

Sry c=22 d=25

Hi Kiran,
We cannot make the denominator = 0, as the value will be undefined and the value has to be finite and defined so to maximize the given expression in question, we have to minimize the denominator & minimum value of denominator is 1.
Hope this helps!

Hi Siva Ram V,
To find the total number of even integers that lie between 1000 and 7000 and have four different digits.
We need to consider the thousands, hundreds, tens, and unit digits of these even integers.

Step 1: Choose the Thousands Digit
- The thousand digit can be one of {1, 2, 3, 4, 5, 6} (six choices).
- We have two cases depending on whether the thousand digit is even or odd.

Case 1: Thousands Digit is Even {2, 4, 6}
- For this case, you have 3 choices for the thousands digit.
- Now, you need to choose the unit digit. Since it's an even number, you have 4 choices (0, 2, 4, 6).
- For the tens and hundreds place, you need to choose from the remaining 8 digits (0, 1, 3, 5, 7, 8, and 9) for the tens place and 7 remaining choices for the hundreds place.
- So, for this case, the total number of even integers = 3 (choices for thousands digit) * 4 (choices for units digit) * 8 (choices for tens place) * 7 (choices for hundreds place) = 3 * 4 * 8 * 7 = 672.

Case 2: Thousands Digit is Odd {1, 3, 5}
- For this case, you also have 3 choices for the thousands digit.
- The unit digit, being even, can be chosen from 5 options (0, 2, 4, 6, 8).
- The tens and hundreds place can still be filled with the remaining 8 and 7 choices, respectively.
- So, for this case, the total number of even integers = 3 (choices for thousands digit) * 5 (choices for units digit) * 8 (choices for tens place) * 7 (choices for hundreds place) = 3 * 5 * 8 * 7 = 840.

Total = Case 1 + Case 2 = 672 + 840 = 1512.

So, there are 1512 even integers between 1000 and 7000 with four different digits.
I hope this helps!