How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Correct Answer: 43200

Explanation:

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

Correct Answer: 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in  10 C 6 or 210 ways. 

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

Correct Answer: 63

Explanation:

Required number of ways =  ( 7 C 5 * 3 C 2 ) = ( 7 C 2 * 3 C 1 ) = 63

Correct Answer: 720

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Correct Answer: 3

Explanation:

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

 

AB , AC

 

BA , BC

 

CA , CB

 

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

 

The lines are: AB, BC and AC ; 3 lines only.

 

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

Correct Answer: 671

Explanation:

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

Correct Answer: 535

Explanation:

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

 

 

 

 

 

 

 

  

Correct Answer: 18!(6 x 24)

Explanation:

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

 

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

Correct Answer: 360

Explanation:

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

Correct Answer: 1092

Explanation:

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