Required number of ways =  8 C 5 * 10 C 6 =   8 C 3 * 10 C 4 = 11760

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6

Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

Required Number of ways = 4 C 2 × 3 C 2 × 2 C 1 = 36 = 36

Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

As n>0, n will not be -8. Therefore, n=11.

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x  P 3 4 = 3 x 24 = 72. 

5 digited numbers =  P 5 5 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Let x, y, z be the number of balls received by the three persons, then

x ? 5 , y ? 5 , z ? 5 a n d x + y + z = 21

Let  u ? 0 , v ? 0 , w ? 0 t h e n

? x + y + z =21

? u + 5 + v + 5 + w + 5 = 21

? u + v + w = 6 

?Total number of solutions =  C 3 - 1 6 + 3 - 1 = C 2 8 = 28

Using,  C r n = C n - r n we get 

n ? 10 = 12

or, n = 12 + 10 = 22