Question 1

Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel ?from A to E?

Accepted Answer

The bus fromA to B can be selected in 3 ways. The bus from B to C can be selected in 4 ways. The bus from C toD can be selected in 2 ways. The bus fromD to E can be selected in 3 ways. So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

Question 2

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Accepted Answer

Required number of ways= 8 C 5 × 10 C 6 = ( 8 C 3 × 10 C 4 )= 11760.

Question 3

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

Accepted Answer

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8!/(2! x 2!)= 10080. Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters =4!/2!= 12. Required number of words = (10080 x 12) = 120960

Question 4

Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct

Accepted Answer

Out of 26 alphabets two distinct letters can be chosen in 26 P 2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. Combined with letters there are 6 P 2 X 100 ways = 65000 ways to choose vehicle numbers.

Question 5

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

Accepted Answer

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12 C 4 = 495. Therefore, we can draw 495 quadrilaterals

Question 6

Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter?

Accepted Answer

Since order does not matter, use the combination formula C 3 4 = 24/6 = 4

Question 7

How many ways are there to deal a five-card hand consisting of three eight's and two sevens?

Accepted Answer

If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed. The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards. We have 4 eights and 4 sevens. We want 3 eights and 2 sevens. C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens) C(4,3) x C(4,2) = 24 Therefore there are 24 different ways in which to deal the desired hand.

Question 8

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

Accepted Answer

ABACUS is a 6 letter word with 3 of the letters being vowels. If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together. These 4 elements can be rearranged in 4! Ways. The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice. Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

Question 9

In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?

Accepted Answer

There are seven positions to be filled. The first position can be filled using any of the 7 letters contained in PROBLEM. The second position can be filled by the remaining 6 letters as the letters should not repeat. The third position can be filled by the remaining 5 letters only and so on. Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

Question 10

There are fourteen juniors and twenty-three seniors in the Service Club. The club is to send four representatives to the State Conference. If the members of the club decide to send two juniors and two seniors, how many different groupings are possib…

Accepted Answer

Choose 2 juniors and 2 seniors. 14 C 2 * 23 C 2 = 23023

Question 11

From a deck of 52 cards, a 5 card hand is dealt.How many distinct hands can be formed if there are atleast 2 queens?

Accepted Answer

The total possible cases would be a 5 card hand with no restrictions : 52 C 5 5 The unwanted cases are: no queens(out of 48 non-queens cards we get 5) 48 C 5 only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4) 4 C 1 * 48 C 4 Therefore, 52 C 5 - ( 48 C 5 + 4 C 1 * 48 C 4 ) = 108336

Question 12

A student council of 5 members is to be formed from a selection pool of 6 boys and 8 girls.How many councils can have Jason on the council?

Accepted Answer

If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4. So, 13 C 4 = 715

Question 13

How many arrangements of the word TRIGONAL can be made if only two vowels and three consonants are used?

Accepted Answer

First we need to choose two vowels 3 C 2 and then three consonants 5 C 3 . Now that we have 5 letters required to make the word,arrange them in 5! ways. So, 3 C 2 * 5 C 3 * 5 ! = 3600

Question 14

A research team of 6 people is to be formed from 10 chemists,5 politicians, 8 economists and 15 biologists.How many teams have atleast 5 chemists?

Accepted Answer

10 C 5 * 28 C 1 * 10 C 6 = 7266

Question 15

There are 7 men and 10 women on a committee selection pool.A committee consisting of President,Vice-President, and Treasurer is to be formed.How many ways can exactly two men be on the committe ?

Accepted Answer

There are 7 C 2 ways of selecting two men, and 10 C 1 ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways. So, 7 C 2 * 10 C 1 * 3 ! = 1260

Question 16

If each of the 8 teams in a league must play each other three times,how many games will be played?

Accepted Answer

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8 C 2 games.Since,each pairing of teams will occur three times,the answer will be triple. Therfore, 8 C 2 * 3 =84

Question 17

In how many different ways can five persons stand in a line for a group photograph?

Accepted Answer

This is the number of permutations of five things taken all at a time. Therefore, answer = 5 P 5 = 120 ways

Question 18

First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?

Accepted Answer

13 P 3 = 13!/10! = 1716

Question 19

When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?

Accepted Answer

He can arrange his schedule in 8 P 6 = 20160 ways

Question 20

A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?

Accepted Answer

The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are: One house gets the wrong package or two houses get the wrong package or three houses get the wrong package. We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package. 1×1×1=1 Determine the total number of ways the three packages can be delivered. 3×2×1=6 The number of ways at least one house gets the wrong package is: 6−1=5 Therefore there are 5 ways for at least one house to get the wrong package.

Permutation and Combination Questions