When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

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: 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: 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

Correct Answer: 25

Explanation:

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.