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Real Life Situations
100

!0=

1

100

Conceptually, what is the difference between permutation and combination

Permutation is a method of arranging all the members in order. Combination is the selection of elements from a collection with no regards to the order they're in.

100

In how many ways can the letters of the word 'LEADER' be arranged?

360

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D, and 1R.

 Required number of ways =6!/(1!)(2!)(1!)(1!)(1!)= 360

100

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

11760

(8C5 x 10C6)

100

A student council has 24 members.  In how many different ways can a president, vice-president, treasurer, historian, and secretary be selected?  

5,100,480

200

The formula for nPr 

n!/(n-r)!

200

Permutation, Combination, or neither:

The batting order for seven players on a 12 person team.

Permutation

200

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

120960

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. And we have MTHMTCS (AEAI).


200

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

210

200

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

756

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

 Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)


300

The formula for nCr

n!/r!(n-r)!

300

We have n objects in total, out of which p are of one type, q are of another type, r are of any other type, remaining objects are all different from each other, 

Find a formula for the total number of ways of arranging them

n!/p!q!r!

300

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

20

(1 x 5 x 4) = 20.

300

On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any one of these games is 2/3.

Find an expression for the probability Alfred wins 4 games on the first day and 2 on the second.

(6C2)2 x (2/3)6 x (1/3)6

300

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

209

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

 Required number of ways= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)


400

If there are x different ways to do a work and y different ways to do another work and both the works are independent of each other, 

then there are _____ ways to do either first OR second work

x+y

400

If we want to arrange n objects in r different spots, and we are free to REPEAT objects as many times as we wish, 

then find the total number of ways of doing this or the total number of permutations

nr

400

You need to put your reindeer, Quentin, Jebediah, Lancer, and Gloopin, in a single-file line to pull your sleigh. However, Gloopin and Lancer are fighting, so you have to keep them apart, or they won't fly.

12

4•3•2•1[total number of arrangements]-(3•2•1)•2[arrangements with Gloopin & Lancer together)

=24-12

=12

400

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

64

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

 Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)=64


400

Eight runners compete in a race where there are no tied finishes. Andrea and Jack are two of the eight competitors in this race.

Find the total number of possible ways in which the eight runners can finish if Jack finishes in any position after Andrea.

20160


[(7C1)+(6C1)+ ... +(1C1)] x 6! = 20160

500

If there are x different ways to do a work and y different ways to do another work and both the works are independent of each other,

then there are ____ ways to do both first AND second works.

x*y

500

A number of distinct points are marked on the circumference of a circle, forming a polygon. Diagonals are drawn by joining all pairs of non-adjacent points. Find an expression for the number of possible diagonals if there are n points, where n>2.

[n x (n-3)]/2

500

A permutation  of  is heavy-tailed if . What is the number of heavy-tailed permutations?

48


There are  total permutations.

For every permutation  such that , there is exactly one permutation such that . Thus it suffices to count the permutations such that .

, , and  are the only combinations of numbers that can satisfy .

There are  combinations of numbers,  possibilities of which side of the equation is  and which side is , and  possibilities for rearranging  and . Thus, there are  permutations such that .

Thus, the number of heavy-tailed permutations is (120-24)/2

500

A fancy bed and breakfast inn has 5 rooms, each with a distinctive color-coded decor. One day 5 friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than 2 friends per room. In how many can the innkeeper assign the guests to the rooms?

2220


The innkeeper can organize the guests into rooms with the following unordered frequencies:

(1,1,1,1,1) (1,1,1,2) or (1,2,2)

  • Case {1,1,1,1,1}: there are 5! ways to assign the guests.
  • Case {1,1,1,2}: there are 5×4 ways to assign the frequencies to rooms, and (5C2)3! ways to assign the guests to rooms (once the frequencies have been assigned).
  • Case {1,2,2}: there are 5×(4C2) ways to assign the frequencies to rooms, and (5C2)(3C2) ways to assign the guests to rooms (once the frequencies have been assigned).

In total this gives:

5!+5×4×(5C2)3!+5×(4C2)×(5C2)(3C2)=2220.


500

Sophie and Ella play a game. They each have five cards showing Roman numerals I, V, X, L, C. Sophie lays her cards face up on the table in order I, V, X, L, C. 

Ella shuffles her cards and lays them face down on the table. She then turns them over one by one to see if her card matches with Sophie's card directly above. Sophie wins if no matches occur; otherwise Ella wins.

I    V   X   L   C    (Sophie's cards)

O   O  O   O   O    (Ella's cards, O=unknown)

Find the probability that Sophie wins the game.


11/30


120-(45+20+10+0+1)=120-76=44

44/120=11/30