Rank of the word ‘GTWENTY’ in dictionary is

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    (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

Similar Questions for you

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

A
alok kumar singh

18 = 32 * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

 

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