The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2, 3, 4, 5, 6 (repetition of digits is not allowed) and divisible by 55 is…………

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    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    1012 Number in the question 23421

    Also, the number has to use digits {2, 3, 4, 5, 6} without repetition and the number has to be divisible by 5 5 1 1 × 5  

    As the number has to be divisible by both 5 and 11,

    5->once place

    Let us make 4-digit such numbers first:

    {2, 3, 4, 6} (digits are not be repeated)

    A number is divisible by 11 it difference of sum of its digits at even places and sum of digits at odd place is 0 or multiple of 11.

    ->Total 6 numbers 3245, 4235, 6325, 2365, 3465, 6435

    Let us make 5 digit such numbers

    2            4       &nb

    ...more

Similar Questions for you

A
alok kumar singh

Start with

(1) E ¯ : 6 ! 2 ! = 3 6 0  

(2)    G E ¯ : 5 ! 2 ! , G N ¯ : 5 ! 2 !  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

A
alok kumar singh

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

A
alok kumar singh

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  5!4!=5

3 2 0 5 ! 3 ! 2 ! = 1 0

3 1 0 5 ! 2 ! 2 ! 2 ! = 1 5

2 1 5 ! 2 ! × 3 ! = 1 0

51 Total way

A
alok kumar singh

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
15+3–1C2 = 17C2 ways

  = 1 7 * 1 6 2 = 1 3 6          

A
alok kumar singh

( n ? 1 ) ( n ? 2 ) ( n ? 3 ) n ( n ? 1 ) ( n ? 2 ) ( n ? 3 ) = 1 8
  n = 8

= 8 * 7 * 6 * 5 * 4 + 9 * 8 2

= 6756

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