In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

Correct Answer: 66

Explanation:

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

 

So, 12 C 21= 66

Correct Answer: 601

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

 

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

 

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

 

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Correct Answer: 186

Explanation:

When at least 2 women are included.

 

The committee may consist of 3 women, 2 men : It can be done in   4 C 3 * 6 C 2  ways

 

or, 4 women, 1 man : It can be done in   4 C 4 * 6 C 1ways

 

or, 2 women, 3 men : It can be done in 4 C 2 * 6 C 3 ways.

 

Total number of ways of forming the committees

 

=  4 C 2 * 6 C 3 + 4 C 3 * 6 C 2 + 4 C 4 * 6 C 1

 

= 6 x 20 + 4 x 15 + 1x 6

 

= 120 + 60 + 6 =186

Correct Answer: 376

Explanation:

The smallest number in the series is 1000, a 4-digit number.

 

The largest number in the series is 4000, the only 4-digit number to start with 4. 

 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

 

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

 

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

 

Including 4000, there will be 376 such numbers.

Correct Answer: 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ( 7 C 3* 4 C 2) 

= 210. 

 

Number of groups, each having 3 consonants and 2 vowels = 210. 

 

Each group contains 5 letters. 

 

Number of ways of arranging 5 letters among themselves = 5! = 120 

 

Required number of ways = (210 x 120) = 25200.

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 '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: 63

Explanation:

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

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: 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