Answer: Option B. -> 71
Case A: Five 6 and one 'zero' = $$\frac{{6!}}{{5!}}$$  = 6
Case B: Four 6 and one '2' and one '4' = $$\frac{{6!}}{{4!}}$$ = 30
Case C: Three 6 and three '4' = $$\frac{{6!}}{{3! \times 3!}}$$  = 20
Case D: Four 6 and two '3' = $$\frac{{6!}}{{4! \times 2!}}$$  = 15
Total number of different sequences = 71

Answer: Option D. -> None of these
The word MANAGEMENT is a 10 letter word.
Normally, any 10 letter word can be rearranged in 10! ways.
However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.
Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to:
$$\frac{{10!}}{{2! \times 2! \times 2! \times 2!}}$$
The problem requires us to find out the number of outcomes in which the two As do not appear together.
The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter.
Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in: $$\frac{{9!}}{{2! \times 2! \times 2!}}$$   ways.
Therefore, the required answer in which the two As do not appear next to each other

Answer: Option D. -> 1701
Number of ways a pair of shoes can be selected,
= 3C1
= 3 ways
Number of ways lower wear can be selected = (3 + 4) = 7 ways.
Number of ways upper wear can be selected,
= 3 + 6 + (3 × 6)
= 27 ways
Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st jacket, 2nd jacket}.
Total number of different outfits
= 3 × 7 × 27 × 3
= 1701 ways

Answer: Option C. -> 252
numbers between 100 and 1000 = 900
Numbers between 100 and 1000 which do not have digit 6 in any place,
= 8 × 9 × 9
= 648
Unit digit could take any value of the 9 values (0 to 9, except 6)
Tens Digit could take any value of the 9 values (0 to 9, except 6)
Hundreds digit could take any value of the 8 values (1 to 9, except 6)
numbers between 100 and 1000 which have at least one digit as 6,
= 900 - 648
= 252

Answer: Option C. -> 540
As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways.
If the first digit is 6, the other digits can be arranged in $$\frac{{6!}}{{2!}}$$ = 360 ways.
If the first digit is 7, the other digits can be arranged in $$\frac{{6!}}{{2! \times 2!}}$$  = 180 ways.
Thus required possibilities for n,
= 360 + 180
= 540 ways

Answer: Option D. -> 200
Number of ways of choosing 6 from 10 = 10C6 = 210
Number of ways of attempting more than 4 from a group,
= 2 × 5C5 × 5C1
= 10
Required number of ways
= 210 - 10
= 200

Answer: Option D. -> 10
If n is even, then the number of boys should be equal to number of girls, let each be a.
⇒ n = 2a
Then the number of arrangements = 2 × a! × a!
If one more students is added, then number of arrangements,
= a! × (a + 1)!
But this is 200% more than the earlier
⇒ 3 × (2 × a! × a!) = a! × (a + 1)!
⇒ a + 1 = 6 and a = 5
⇒ n = 10
But if n is odd, then number of arrangements,
= a!(a + 1)!
Where, n = 2a + 1
When one student is included, number of arrangements,
= 2(a + 1)! (a + 1)!
By the given condition, 2(a + 1) = 3, which is not possible.

Answer: Option C. -> 216
Test of divisibility for 3:
The sum of the digits of any number that is divisible by 3 is divisible by 3
For instance, take the number 54372
Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21
As 21 is divisible by 3, 54372 is also divisible by 3
There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by 3
Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either 0 or 3 while forming the five digit numbers.
Case 1
If we do not use '0', then the remaining 5 digits can be arranged in:
5! ways = 120 numbers.
Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.
The first digit from the left can be any of the 4 digits 1, 2, 4 or 5
Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.
So, there will be 4 × 4! numbers = 4 × 24 = 96 numbers.
Combining Case 1 and Case 2, there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.

Answer: Option D. -> 376
The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 3 × 5 × 5 × 5 or 375 numbers from 1000 to 3999
Including 4000, there will be 376 such numbers.

Answer: Option B. -> 24
Any factor of this number should be of the form 2a × 3b × 5c
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4, b can take values 0, 2, 4, 6 and c can take values 0, 2
Total number of perfect squares =3 × 4 × 2 = 24