JEE Main Maths Test- 12 - JEE MCQ

Clearly, |z−1|=1 represents a circle with centre at (1,0) and radius 1.
Let P(z) be any point on it.

Then, arg(z−1) =∠XCP = θ [say]
Therefore, arg(z) =∠XOP = θ/2
Hence, arg(z−1) = 2arg(z)

The lady has ten friends. Two of these friends will not agree to go to the party together. So, let us consider the following case: neither of the two friends will attend the party. So, the number of ways six guests can be invited in this case will be equal to the number of ways to choose 6 friends out of the remaining 8 friends. So, we have

Therefore, there are 28 ways to invite 6 friends when the two friends will not be going to the party.
Now, we will consider the next case, which is that one of the two friends decides to attend the party. So, if either of the two decide to attend the party, then the rest 5 guests will be chosen from the remaining 8 friends. Therefore, if one friend decides to attend the party, the number of ways to invite the other 5 guests is ⁸C₅ . Similarly, if the other friend decides to attend the party, the number of ways to invite the other 5 guests is ⁸C₅ . Therefore, the number of ways to invite the guests if either, if the two friends is going to the party, is the following,

Therefore, the total number of ways to invite six guests out of ten with the given conditions is 28+112 = 140.

To solve this problem, we need to calculate the sum of the digits at the unit place of all numbers that can be formed using the digits 3, 4, 5, and 6, taken all at a time.

Step 1: Find the total number of numbers formed

Since we are arranging 4 digits (3, 4, 5, 6) in all possible ways, the total number of numbers formed is 4!4!4! (4 factorial), which equals:

4!=4×3×2×1=24

So, 24 numbers are formed.

Step 2: Consider the unit place

Each digit (3, 4, 5, 6) will appear at the unit place equally in all possible permutations. Since there are 24 numbers and 4 digits, each digit will appear 24/4=6 times in the unit place.

Step 3: Calculate the sum of digits at the unit place

Now, we calculate the sum of the digits appearing at the unit place:

(3+4+5+6)×6=18×6=108

Thus, the sum of the digits at the unit place of all the numbers formed is 108.

The correct answer is:

2. 108.

Step 1: Identify the letters and their order
The letters in "RANDOM" are A, D, M, N, O, R. Arranging them in alphabetical order gives us: A, D, M, N, O, R.

Step 2: Count the arrangements starting with letters before 'R'
1. Words starting with 'A':
- Remaining letters: D, M, N, O, R
- Number of arrangements = 5! = 120

2. Words starting with 'D':
- Remaining letters: A, M, N, O, R
- Number of arrangements = 5! = 120

3. Words starting with 'M':
- Remaining letters: A, D, N, O, R
- Number of arrangements = 5! = 120

4. Words starting with 'N':
- Remaining letters: A, D, M, O, R
- Number of arrangements = 5! = 120

5. Words starting with 'O':
- Remaining letters: A, D, M, N, R
- Number of arrangements = 5! = 120

Step 3: Count the arrangements starting with 'R'
Now we focus on words starting with 'R'.

1. Fix 'R' and consider the next letter:
- The next letter must be 'A' (the only letter before D, M, N, O):
- Remaining letters: D, M, N, O

2. Fix 'RA' and consider the next letter:
- The next letter can be 'D':
- Remaining letters: M, N, O
- Number of arrangements = 3! = 6

3. Fix 'RAD' and consider the next letter:
- The next letter can be 'M':
- Remaining letters: N, O
- Number of arrangements = 2! = 2

4. Fix 'RAM' and consider the next letter:
- The next letter can be 'N':
- Remaining letters: O
- Number of arrangements = 1! = 1

5. Fix 'RAN' and consider the next letter:
- The next letter must be 'D':
- Remaining letters: O, M
- Number of arrangements = 2! = 2

6. Fix 'RAND' and consider the next letter:
- The next letter must be 'O':
- Remaining letter: M
- Number of arrangements = 1! = 1

Step 4: Total the arrangements
Now we can sum all the arrangements:
- Words starting with 'A': 120
- Words starting with 'D': 120
- Words starting with 'M': 120
- Words starting with 'N': 120
- Words starting with 'O': 120
- Words starting with 'RA': 6
- Words starting with 'RAD': 2
- Words starting with 'RAM': 1
- Words starting with 'RAN': 2
- Words starting with 'RAND': 1

Total = 120 + 120 + 120 + 120 + 120 + 6 + 2 + 1 + 2 + 1 = 614

Conclusion
The rank of the word "RANDOM" is 614.

Hence there are four such numbers.