The divisibility rule of 3 says that a given number will be divisible by 3, if the sum of the digits is divisible by 3. Now, the sum of the given seven digits is 21.

Since, we need to form 6 digit numbers; hence one of the digits will be left out while forming the desired numbers. Hence, the only possible cases are when 0, 3 and 6 are left one at a time.

1. When 0 is left out, the total numbers would be 6!.
2. When 3 is left out. The number of ways of numbers that can be formed = 5 × 5 × 4 × 3 × 2 × 1 = 600. Thus, a total of 5 × 5! Numbers can be formed.
3. When 6 is left out, again a total of 5 × 5! Numbers can be formed.

So, Total numbers formed =2 × 5 × 5! + 6! = 1920

Test of divisibility of 11: (Sum of digits at odd places ~ Sum of digits even digits) is divisible by 11 or equal to zero.

Given number of digit = 3, 4, 5, 6, 7

(Odd)  (Even)  (Odd)  (Even)  (Odd)

Now two digits will be at even places.

Pair of Numbers at even places - (3,4); (3,5); (3,6); (3,7); (4,5); (4,6); (4,7); (5,6); (5,7); (6,7).

Their respective sums are 7, 8, 9, 10, 9, 10, 11, 11, 12, and 13.

And the sum of other three digits will be 18, 17, 16, 15, 16, 15, 14, 14, 13, and 12.

Hence the differences will be 11, 9, 7, 5, 7, 5, 3, 3, 1, and 1. So there is only one case of difference 11 i.e. (Sum of odd digits from) – (Sum of even digits)

=> (5 + 6 + 7) – (3 + 4) = 11

Digits at odd places (5, 6, 7) can be arranged in 3P3 = 3! = 6 ways.

Digits at even places (3, 4) can be arranged in 2P2 = 2! = 2 ways.

Total number of numbers which are multiples of 11 are = 6 × 2 = 12

Here odd and even positions are:

 C    O   M    P    U    T   E   R

(O) (E) (O) (E) (O) (E) (O) (E)

There are 3 vowels O, U , E and 5 consonants.

Since vowels are to be arranged only at odd positions and there are 4 odd positions. So, number of arrangements for vowels = 4P3 = 24

And remaining 5 consonants C, M, P, T and R can be arranged at 5 places in 5! = 120 ways.

So, total number of ways = 120 x 24 = 2800

The number of arrangements = 7P4 = 7 x 6 x 5 x 4 = 840

Word ENGINEERING has 11 letters, from which EIEEI are vowels, and NGNRNG are consonants.

So, number of arrangements for vowels = 5! / (2! X 3!) = 10

Number of arrangements of 7 objects with consonants (NGNRNG) = 7! / (2! X 3!) = 420

Total number of arrangements = 420 x 10 = 4200 (by rule of multiplication)