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 Permutations and Combinations Questions

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FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

 

 

1.  Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.

Examples : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

 

2.  Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

Prn=nn-1n-2....n-r+1=n!n-r!

 

Ex : (i) P26=6×5=30   (ii) P37=7×6×5=210

Cor. number of all permutations of n things, taken all at a time = n!.

Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that p1+p2+...+pr=n

Then, number of permutations of these n objects is :

n!(p1!)×(p2! ).... (pr!)

 

3.  Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Crn=n!(r !)(n-r)!=nn-1n-2....to r factorsr!

 

Note : (i)Cnn=1 and C0n =1     (ii)Crn=C(n-r)n

 

Examples : (i) C411=11×10×9×84×3×2×1=330      (ii)C1316=C(16-13)16=C316=560

Q:

How many four digit even numbers can be formed using the digits {2, 7, 5, 3, 9, 1} ?

A) 59 B) 60
C) 61 D) 64
 
Answer & Explanation Answer: B) 60

Explanation:

The given digits are 1, 2, 3, 5, 7, 9  

A number is even when its units digit is even. Of the given digits, two is the only even digit.Units place is filled with only '2' and the remaining three places can be filled in ⁵P₃ ways.

 

Number of even numbers = ⁵P₃ = 60.

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Q:

If a+b+c = 21, what is the total number of non - negative integral solutions?

A) 123 B) 253
C) 321 D) 231
 
Answer & Explanation Answer: B) 253

Explanation:

Number of non - negative integral solutions =n+r-1 Cr-1 = C223 = 253 

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4 7154
Q:

What is the total no of ways of selecting atleast one item from each of the two sets containing 6 different items each?

A) 2856 B) 3969
C) 480 D) None of these
 
Answer & Explanation Answer: B) 3969

Explanation:

We can select atleast one item from 6 different items = 26-1  

 

Similarly we can select atleast one item from other set of 6 different items in 26-1 ways. 

 

Required number of ways = (26-1)26-1 = 26-12 = 3969

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1 7018
Q:

A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ?

A) ²²C₁₀ + 1 B) ²²C₉ + ¹⁰C₁
C) ²²C₁₀ D) ²²C₁₀ - 1
 
Answer & Explanation Answer: D) ²²C₁₀ - 1

Explanation:

The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1

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2 6934
Q:

What is the value of 100P2 ?

A) 10000 B) 9900
C) 8900 D) 7900
 
Answer & Explanation Answer: B) 9900

Explanation:

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

 

That can be done as,

 

100 P2  = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

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7 6914
Q:

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?

A) 3 B) 5
C) 3! D) 5!
 
Answer & Explanation Answer: B) 5

Explanation:

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.

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2 6834
Q:

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

A) 36 B) 72
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

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.

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1 6792
Q:

The letters of the word PROMISE are to be arranged so that three vowels should not come together. Find the number of ways of arrangements?

A) 4320 B) 4694
C) 4957 D) 4871
 
Answer & Explanation Answer: A) 4320

Explanation:

Given Word is PROMISE.

Number of letters in the word PROMISE = 7

Number of ways 7 letters can be arranged = 7! ways

Number of Vowels in word PROMISE = 3 (O, I, E)

Number of ways the vowels can be arranged that 3 Vowels come together = 5! x 3! ways

 

Now, the number of ways of arrangements so that three vowels should not come together

= 7! - (5! x 3!) ways = 5040 - 720 = 4320.

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