What is the number of possible words that can be made using the word quiz such that the vowels never come together Choices 8/12 16 24?

Latest Permutation and Combination MCQ Objective Questions

Permutation and Combination MCQ Question 1:

In how many ways can 6 boys be seated in a row such that two boys [Ashok and Pankaj] are never seated together.

1. 720
2. 240
3. 480
4. 360
5. None of the above/More than one of the above

Answer [Detailed Solution Below]

Option 3 : 480

Concept:-

For these kind of questions we take the number of ways in which they are seated together and subtract them from the total number of ways.

Calculation:-

Number of ways of seating 6 boys in a row = 6! = 720

Number of ways in which Ashok and Pankaj are sitting together can be found by taking both of them as 1.

So number of ways both of them are sitting together = [AP + 4]! + [PA + 4]! = 5! × 2 = 240

Total number of ways = 720 - 240 = 480

Permutation and Combination MCQ Question 2:

In how many ways a 4-digit number greater than 5000 can be formed from the digits 1, 2, 3, 4, 5, 6, 7 and 8 if repetition is allowed.

1. 83 × 5
2. 47 × 4
3. 8C4
4. 83 × 4
5. None of the above/More than one of the above

Answer [Detailed Solution Below]

Option 4 : 83 × 4

GIVEN:

8 digits i.e. 1, 2, 3, 4, 5, 6, 7 and 8.

CONCEPT:

For number to be greater than 5000, 1st digit should be greater than or equal to 5.

CALCULATION:

As, we have to use digit from 1 to 8 and 1st digit can be greater than or equal to 5, so 1st digit can be any of 5, 6, 7, and 8.

⇒ 4 ways of selecting 1st digit.

Number of ways of selecting other digits = 8 each.

⇒ Total number of ways = 4 × 8 × 8 × 8 = 83 × 4 number of ways.

Permutation and Combination MCQ Question 3:

In how many different ways the letters of the word "BANK" can be arranged?

1. 12
2. 24
3. 16
4. 8

Answer [Detailed Solution Below]

Option 2 : 24

Given:

The word "BANK" is given.

⇒ In how many different ways the letters of the word "BANK" can be arranged.

Calculation:

Number of letters in the word = 4

Total ways = 4!

Total ways = 4 × 3 × 2 × 1

⇒ Total ways = 24

Hence, the correct answer is "24".

Permutation and Combination MCQ Question 4:

In how many ways can 5 toys be selected out of 12 toys if a particular toy is always included?

1. 990
2. 330
3. 792
4. 7920

Answer [Detailed Solution Below]

Option 2 : 330

Given:

Total Toys = 12

Toys to be selected = 5

⇒ A particular toy is always included.

Calculation:

So, Remaining toys = 11

Toys to be selected = 4

So, Total ways = 11C4

⇒ 11!/[4! × 7!]

⇒ [11 × 10 × 9 × 8]/[4 × 3 × 2 × 1]

⇒ 330

⇒ Total ways = 330

Hence, the correct answer is "330".

Permutation and Combination MCQ Question 5:

How many numbers can be formed from the nine digits 1, 2, ........ 9 by using the six digits?

1. 72
2. 64840
3. 60480
4. 120

Answer [Detailed Solution Below]

Option 3 : 60480

Total digits = 1 to 9

Number from from digits = 6

Total 6-digit numbers = 9P6

⇒ 9!/3!

⇒ 9 × 8 × 7 × 6 × 5 × 4

⇒ 60480

So, Total 60480 6-digit numbers can be formed.

Hence, the correct answer is "60480".

How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 [repetition of digits is allowed]?

1. 10
2. 9
3. 7
4. 8

Answer [Detailed Solution Below]

Option 2 : 9

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are:

33, 35, 37, 53, 55, 57, 73, 75, 77 

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

1. 2520
2. 2530
3. 15130
4. 15120

Answer [Detailed Solution Below]

Option 4 : 15120

Given:

The given number is 'GEOGRAPHY'

Calculation:

The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY[EOA].

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,

The number of ways to arrange these letters = 7!/2!

⇒ 7 × 6 × 5 × 4 × 3 = 2520

In the 3 vowels[EOA], all vowels are different

The number of ways to arrange these vowels = 3!

⇒ 3 × 2 × 1 = 6

Now, 

The required number of ways = 2520 × 6 

⇒ 15120

∴ The required number of ways is 15120.

How many 3 digit odd numbers can be formed from the digits 5, 6, 7, 8, 9, if the digits can be repeated

1. 55
2. 75
3. 70
4. 85

Answer [Detailed Solution Below]

Option 2 : 75

Given:

5, 6, 7, 8, 9 are the digits to form 3 digit number

Calculation:

Let us take the 3digit number as H T U [Hundreds, tens, unit digit]  respectively

To make 3 digit number as odd

5, 7, 9 are only possibly be used in the unit digit place

In hundreds and tens place all  5 digits are possible 

Number of ways for unit digit = 3

Number of ways for tens digit = 5

Number of ways for hundreds digit = 5

Number of 3 digits odd number =  3 × 5 × 5 = 75 

∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated

In how many ways can the word CHRISTMAS be arranged so that the letters C and M are never adjacent?

1. 8! × [7/2]
2. 9! × [7/2]
3. 8! × [9/2]
4. None of the above

Answer [Detailed Solution Below]

Option 1 : 8! × [7/2]

Given:

Different words from CHRISTMAS have to be formed.

Formula:

Words in which letters C and M are never adjacent = All cases – words having C and M together.

Calculation:

⇒ Total number of words = 9!/2! [Division of 2! As S is repeated]

Let us assume C and M to be one unit. Then, letters can be arranged in 8! Ways. C and M can be arranged in 2! Ways. Letter S is repeated, so total number of ways will be divided by 2!

⇒ Number of words with C and M adjacent= 8!/2! × 2! = 8!

Words in which letters C and M are never adjacent = 9!/2! – 8! = 8! × [9/2 - 1] = 8! × [7/2]

How many four-digit numbers can be formed with digits 2, 5, 6, 7 and 8? [Repeating digits are not allowed]

1. 120
2. 115
3. 110
4. 113

Answer [Detailed Solution Below]

Option 1 : 120

Given:

5 number are given 2, 5, 6, 7 and 8

Four-digit number without repetition

Formula used:

Permutation for no repetition = \[\frac{n!}{[n \ - \ r]!}\]

Where n  = total possible numbers

r = required number

Calculation:

Here the total possible number n = 5

And Required number r = 4

Applying the formula

\[\frac{5!}{[5\ - \ 4]!}\]

⇒ 5!

⇒ 5 × 4 × 3 × 2 × 1 = 120

∴ There will 120 possible four-digit number.

In how many ways can we sort the letters of the word MANAGEMENT so that the comparative position of vowels and consonants remains the same as in MANAGEMENT.

1. 1280
2. 720
3. 960
4. 1080

Answer [Detailed Solution Below]

Option 4 : 1080

Given:

Word = MANAGEMENT

Calculation:

Vowel occupy 4 places then !4

∵ A and E are repeated then !4/[!2 × !2]

Consonant occupy 6 places then !6

⇒ M and N are repeated then !6/[!2 × !2]

 ∴ \[\frac{{!4}}{{!2\; × \;!2}}\; × \;\frac{{!6}}{{!2\; × \;!2}} = \;\frac{{4\; × 3\; × 2\; × 1}}{{2\; × 1\; × 2\; × 1}}\; × \frac{{6\; × 5\; × 4\; × 3\; × 2\; × 1}}{{2\; × 1\; × 2\; × 1}}\]

= 6 × 180 = 1080

How many different 6-digit numbers can be formed from the digits 4, 5, 2, 1, 8, 9 ?

1. 480
2. 360
3. 720
4. 840

Answer [Detailed Solution Below]

Option 3 : 720

The given digits are 4, 5, 2, 1, 8, 9

Therefore required number of ways = \[\frac{n!}{n_1!n_2!...n_k!}\] 

\[\frac{6!}{[1!][1!][1!][1!][1!][1!]}=720\]

Hence, the correct answer is 720.

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

1. 645
2. 564
3. 735
4. 756

Answer [Detailed Solution Below]

Option 4 : 756

Given: 

[7 men + 6 women] 5 persons are to be chosen for a committee.

Formula used:

nCr = n!/[n - r]! r!

Calculation:

Ways in which at least 3 men are selected;

⇒ 3 men + 2 women

⇒ 4 men + 1 woman 

⇒ 5 men + 0 woman 

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0

⇒ 7!/[3! × 4!] × 6!/[2! × 4!] + 7!/[4! × 3!] × 6!/[1! × 5!] + 7!/[5! × 2!] × 6!/[6!× 0!]

⇒ 35 × 15 + 35 × 6 + 21 

⇒ 735 + 21 = 756

∴ The required no of ways = 756.

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

1. 110
2. 120
3. 105
4. 115

Answer [Detailed Solution Below]

Option 2 : 120

Given 

Total alphabets in word 'FIGHT' = 5 

Concept Used 

Total number ways of arrangement = n! 

Calculation 

The number of different ways of arrangement of n different words [without repetition] = 5! 

⇒ 5 × 4 × 3 × 2 × 1 = 120 

∴ The required answer is 120 

How many arrangements can be made using the word TESTBOOK so that all the vowels come together

1. 720
2. 1080
3. 4320
4. 840

Answer [Detailed Solution Below]

Option 2 : 1080

Consider all the vowels together as one unit like TSTBK [EOO]

Now, total number of units  is [5 + 1]  = 6. 

the units can arrange themselves in 6! ways.

the vowels [EOO] can arrange themselves in 3! ways.

Now both letter 'T' and 'O' repeat two times

So, the total number of ways = 6! ×  3!/[2! × 2!] = 4320/4 = 1080

Find the number of ways of the word ANUBHAW can be arranged.

1. 2250
2. 5220
3. 2500
4. 2520

Answer [Detailed Solution Below]

Option 4 : 2520

Concept used:

Number of all permutations of n things = n!

Calculation:

The word ANUBHAW contain 7 letters but the letter A appears two times

⇒ No. of way = 7!/2!

⇒ [7 × 6 × 5 × 4 × 3 × 2!]/2!

⇒ 7 × 6 × 5 × 4 × 3

⇒ 2520

∴ The word ANUBHAW can be rearranged in 2520 ways

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

1. 72
2. 144
3. 360
4. 720

Answer [Detailed Solution Below]

Option 3 : 360

Given

Word = LEADER

Concept

A word having 'n' letters in which 'a' letters are repeated can be written in n!/a! ways.

Calculation

Number of letters in LEADER = 6

Number of repeated letters = 2

∴ Required ways = 6!/2! = 6 × 5 × 4 × 3 = 360

A team of 3 doctors has to be selected having at least one man. If there are a total of 4 men and 3 women, in how many ways the team can be formed?

1. 30
2. 34
3. 24
4. 32
5. 20

Answer [Detailed Solution Below]

Option 2 : 34

In the team, there can be one man, two men or three men.

Number of ways of selecting one man and two women = 4C1 × 3C2 = 4 × 3 = 12

Number of ways of selecting two men and one woman = 4C2 × 3C1 = 6 × 3 = 18

Number of ways of selecting three men = 4C3 = 4

∴ Total number of ways = 12 + 18 + 4 = 34

Alternate Method

Number of ways of selecting at least one man in the team = total ways of selecting - number of ways of selecting only women

⇒ 7C3 - 3C3

⇒ 35 - 1

⇒ 34

In how many different ways the letters of the word ‘MENHIR’ be arranged so that the vowel never comes together?

1. 720
2. 240
3. 480
4. 960

Answer [Detailed Solution Below]

Option 3 : 480

Given:

The given word = ‘MENHIR’

Concept used:

Assume all the vowels equal to one letter

Formula used:

If a word has n letters, then the number of different ways to arrange the letter is

Case1:- When no repetition of letters takes place = n!

Case2:- When r1, r2, r3, … rn repeated letters = n!/[r1! r2! … rn!]

Calculation:

The given word = ‘MENHIR’

Total number of letters in the given word = 6

The total number of ways to arrange ‘MENHIR’ = 6! = 720

Number of vowel = 2

Total number of letters in the given word after taking two vowels as one = 5

The number of ways to arrange the given word = 5!

The vowel can be arranged in 2! Ways

Total number of ways to arrange the word when the vowel is together = 5! × 2! = 240

Total number of ways = 720 – 240 = 480

∴ The number of ways to arrange the given word when a vowel is never together is 480

In how many ways can three men and five boys be seated in a linear arrangement so that all the men sit together?

1. 5000
2. 4200
3. 4800
4. 4320

Answer [Detailed Solution Below]

Option 4 : 4320

Given:

Three men and five boys are to be seated in a linear fashion so that men sit together.

Formula:

Treat the persons who sit together as one and arrange. Then, arrange the group internally.  

Calculation:

Men can be treated as one group.

So, five boys and 1 group of men can be arranged in 6! ways.

Three men can be arranged in 3! ways.

Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways

∴ Required no. of ways = 4320

A box contains 5 red, 4 white and 3 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours.

1. 24
2. 32
3. 60
4. 30

Answer [Detailed Solution Below]

Option 3 : 60

1 red ball can be selected in 5C1 ways

1 white ball can be selected in 4C1 ways

1 blue ball can be selected in 3C1 ways

∴ Total number of ways =  5C1 × 4C1 × 3C1 = 5 × 4 × 3 = 60

A box contains 4 red balls, 7 green balls and 5 yellow balls. In how many ways 4 balls can be drawn from the box?

1. 1520
2. 1620
3. 1720
4. 1820

Answer [Detailed Solution Below]

Option 4 : 1820

Given:

Total number of balls = [4 + 7 + 5] = 16

Concept used:

When any specified colour of the ball is not mentioned at the time of drawing the ball then we can pick any ball from the box.

Formula used:

The different combinations of ‘n’ distinct objects taken ‘r' at a time are,

C[n, r] = n!/{r! [n – r]!}

Calculation:

Total number of balls, [n] = 16

Number of balls to be drawn, [r] = 4

Number of ways to draw balls = C[16, 4]

⇒ 16!/[12! 4!]

⇒ [16 × 15 × 14 × 13]/[4 × 3 × 2]

⇒ 1820

∴ The number of ways to draw balls from the box is 1820

From a group of 8 men and 7 women, in how many ways can 6 men and 4 women be selected?

1. 740
2. 840
3. 754
4. 980

Answer [Detailed Solution Below]

Option 4 : 980

Given:

Group of 8 men and 7 women.

Formula used:

\[{{\rm{n}}_{{{\rm{C}}_{r\;}} = \;\frac{{{\rm{n}}!}}{{{\rm{r}}!{\rm{\;}}\left[ {{\rm{n\;}} - {\rm{\;r}}} \right]!{\rm{\;}}}}}}\]

Calculation:

8C6× 7C4

out of total 8 men, 6 men selected = 8C6

out of total 7 women, 4 women selected = 7C4

8C6 × 7C4

\[\Rightarrow \;\frac{{8!}}{{6!\; × \;2!}}\; × \;\frac{{7!}}{{4!\; × \;3!}}\]

\[\Rightarrow \;\frac{{8\; × \;7\; × \;6!}}{{6!\; × \;2}}\; × \;\frac{{7\; × \;6\; × \;5\; × \;4!}}{{4!\; × \;3\; × 2}}\]

⇒ 28 × 35

⇒ 980

∴ 6 men and 4 women are selected in 980 ways.

How many words can be formed from the letters of the word ‘DESIGN’ such that the both vowels are at two ends of the word?  

1. 24
2. 36
3. 48
4. 72

Answer [Detailed Solution Below]

Option 3 : 48

GIVEN:

Word: ‘DESIGN’

CONCEPT:

Basic concept of Permutation and combination.

Number of ways in which ‘x’ letters can be placed in ‘x’ places = x!

CALCULATION:

Total letters in the word ‘DESIGN’ = 6

Vowels: E and I

Since, we want that the vowels must be at two ends of the word, they can be shuffled in itself only.

∴ Required number of words = [6 – 2]! × 2! = 4! × 2! = 48

At the end of a business conference the ten people present all shake hands with each other once. How many handshakes will there be altogether?

1. 20
2. 45
3. 55
4. 90

Answer [Detailed Solution Below]

Option 2 : 45

Given:

At the end of a business conference, the ten people present all shake hands with each other once.

Formula:

Number of Combinations: 

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

nCr = n! / [[r!] [n – r]!]

Calculation:

Required number of shake hands = 10C2 = [10 × 9] / [2 × 1] = 45.

What is the number of possible words that can be made using the word quiz such that was never come together?

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How many words of 3 consonants and 2 vowels can be made from 12 consonants and 4 vowels if all the letters are different?

How many ways a 6 member team can be formed having 3 men and 3 ladies?