Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?

Correct Answer: 126

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

 

We may divide the 8 students as follows

 

Case I: 5 students in the first car and 3 in the second Or

 

Case II: 4 students in the first car and 4 in the second

 

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in  8 C 3 ways.

 

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in  8 C 4 ways.

 

Therefore, the total number of ways in which 8 students can travel is

 

8 C 3 + 8 C 4 = 56 + 70 = 126.

Correct Answer: 300

Explanation:

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

 

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

 

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

 

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

 

In each of these numbers, 7 is written once. Therefore, 243 times.

 

 

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

 

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

 

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

 

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

 

 

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

 

Therefore, the total number of times the digit 7 is written between 1 and 999 is

 

243 + 54 + 3 = 300

Correct Answer: 1

Explanation:

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

Correct Answer: 1024

Explanation:

There are 5 slots.

 

                   __ __ __ __ __

 

The first slot must be a four. There are 4 ways to put a four in the first slot.

 

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

 

(4)(4)(4)(4)(4) = 1024

 

Therefore there are 1024 different ways to produce the desired hand of cards.

Correct Answer: 30

Explanation:

Since order does not matter it is a combination. 

 

The word AND means multiply. 

 

Given 4 basketball, 3 volleyball, 2 soccer. 

 

We want 2 basketball games and 1 other event. There are 5 choices left. 

C(n,r) 

C(How many do you have, How many do you want) 

C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) 

C(4,2) x C(5,1) = (6)(5) = 30

 

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.

Correct Answer: 2 x 18!

Explanation:

fix one person and the brothers B1 P B2 = 2 ways to do so.
other 17 people= 17!

 
Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!

Correct Answer: 12600

Explanation:

Total number of alphabets = 10

so ways to arrange them = 10! 

 

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps 

 

Hence number of arrangements = 10!/4! x 3! x 2! = 12600

Correct Answer: 6

Explanation:

Selecting zero'A's= 1

Selecting one 'A's = 1

Selecting two 'A's = 1

Selecting three 'A's = 1

Selecting four 'A's = 1

Selecting five 'A's = 1

=> Required number ofways =6

Correct Answer: 360

Explanation:

Two horses A and B, in a race of 6 horses... A has to finish before B

 

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

 

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

 

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

 

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

 

A cannot finish 6th, since he has to be ahead of B

 

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

Correct Answer: 126

Explanation:

This question is a combination since order is not important.

 

Answer =  7 C 3 = 126