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:

$P_{r}^{n} = n (n - 1) (n - 2) . . . . (n - r + 1) = \frac{n!}{(n - r)!}$

Ex : (i) $P_{2}^{6} = (6 \times 5) = 30$ (ii) $P_{3}^{7} = (7 \times 6 \times 5) = 210$

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 $(p_{1} + p_{2} + . . . + p_{r}) = n$

Then, number of permutations of these n objects is :

$\frac{n!}{(p_{1}!) \times (p_{2}!) . . . . (p_{r}!)}$

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:

$C_{r}^{n} = \frac{n!}{(r!) (n - r)!} = \frac{n (n - 1) (n - 2) . . . . t o r f a c t o r s}{r!}$

Note : (i) $C_{n}^{n} = 1 a n d C_{0}^{n} = 1$ (ii) $C_{r}^{n} = C_{(n - r)}^{n}$

Examples : (i) $C_{4}^{11} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$ (ii) $C_{13}^{16} = C_{(16 - 13)}^{16} = C_{3}^{16} = 560$

How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000?

A) 243	B) 300
C) 301	D) 290

Answer & Explanation Answer: B) 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

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0 6847

The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is ?

A) 2(6!)	B) 6! x 7
C) 6! x ⁷P₆	D) None

Answer & Explanation Answer: C) 6! x ⁷P₆

Explanation:

We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.

Hence required number of ways = 6! x ⁷P₆

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6 6710

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 = $2^{6} - 1$

Similarly we can select atleast one item from other set of 6 different items in $2^{6} - 1$ ways.

Required number of ways = $(2^{6} - 1) (2^{6} - 1) = {(2^{6} - 1)}^{2}$ = 3969

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1 6567

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 6335

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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7 6327

In how many ways the letters of the word OLIVER be arranged so that the vowels in the word always occur in the dictionary order as we move from left to right ?

A) 186	B) 144
C) 136	D) 120

Answer & Explanation Answer: D) 120

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x ${}^{6}C_{3}$ ways = 6x5x4 = 120 ways.

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11 6320

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 6261

When six fairs coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads ?

A) 15	B) 42
C) 16	D) 40

Answer & Explanation Answer: B) 42

Explanation:

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is $6 C_{0}$ =1

The number of outcomes in which 1 coin turns head is = $6 C_{1}$ =6

The number of outcomes in which 2 coins turn heads is $6 C_{2}$ =15

The number of outcomes in which 3 coins turn heads is $6 C_{3}$ =20

Therefore, total number of outcomes =1+6+15+20= 42 outcomes

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