There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged =  9 ! 2 ! × 2 ! × 2 ! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in  6 ! 2 ! × 2 ! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in  4 ! 2 ! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

=  10 P 4

= 5040.

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 ? 625 = 671

The number of points of intersection of 37 lines is  C 2 37. But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting  C 2 13 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting  C 2 11 points, we get only one point B.

 Hence the number of intersection points of the lines is  C 2 37 - C 2 13 - C 2 11 + 2 = 535

Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular?permutations: 18 P 12 2 * 12 =  18 ! 6 * 24

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

 Number of 7 digit numbers =  7 ! 3 ! × 2 ! = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

= 6 ! 3 ! × 2 ! = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360

We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2 C 1 * 5 C 4 * 9 C 6 = 840

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2 C 1 * 5 C 5 * 9 C 5 =252

Total number of ways of selecting the team = 840 + 252 = 1092

The toys are different; The boxes are identical 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

Two toys out of the 5 can be selected in  5 C 2 ways. Another 2 out of the remaining 3 can be selected in  3 C 2 ways and the last toy can be selected in  1 C 1 way. 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5 C 2 * 3 C 2= 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in  5 C 3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is  5 C 3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

The number of ways of arranging n beads in a necklace is  ( n - 1 ) ! 2 = ( 8 - 1 ) ! 2 = 7 ! 2 = 2520 

(since n = 8)