A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

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: 720

Explanation:

The word 'LEADING' has 7 different letters.

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

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.

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

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

Correct Answer: 66

Explanation:

There are 12 people, so this is our n value.

 

So, 12 C 21= 66

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: 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

 

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

 

=  10 P 4

 

= 5040.

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