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Maths Permutations and Combinations Test.8

**Quiz Instructions:**

- There will be 20 multiple choice question in this online test.
- Answer of the questions will change randomly each time you start this test.
- Practice this test at least 3 times if you want to secure High Marks.
- At the End of the Test you can see your Test score and Rating.

1 / 20

**Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by wavingI. Single flag of different coloursII. Any two flags in a different sequence of colours.III. three flags in a different sequence of colours.The maximum number of codes that can be generated is.**

Explanation:

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

We can choose any colour.

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

Explanation:

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

We can choose any colour.

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

2 / 20

**A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?**

Explanation:

Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 45 – 1.

= 210 – 1 = 1024 – 1 = 1023

Explanation:

Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 45 – 1.

= 210 – 1 = 1024 – 1 = 1023

3 / 20

**In a class there are 20 boys and 25 girls. In how many ways can a boy and a girl be selected?**

Explanation:

We can select one boy from 20 boys in 20 ways.

We select one girl from 25 girls in 25 ways

We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

Explanation:

We can select one boy from 20 boys in 20 ways.

We select one girl from 25 girls in 25 ways

We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

4 / 20

**In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?**

Explanation:

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.

3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

Explanation:

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.

3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

5 / 20

**A bag contains nine yellow balls, three white balls and four red balls. In how many ways can two balls be drawn from the bag?**

Explanation:

Total number of balls = 9 + 3 + 4

Two balls can be drawn from 16 balls in ¹⁶C₂ ways.

Explanation:

Total number of balls = 9 + 3 + 4

Two balls can be drawn from 16 balls in ¹⁶C₂ ways.

6 / 20

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

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.

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.

7 / 20

**How many four digit numbers can be formed using the digits {1, 3, 4, 5, 7,9}(repetition of digits is not allowed)?**

Explanation:

The given digits are six.

The number of four digit numbers that can be formed using six digits is ⁶P₄ = 6 * 5 * 4 * 3 = 360.

Explanation:

The given digits are six.

The number of four digit numbers that can be formed using six digits is ⁶P₄ = 6 * 5 * 4 * 3 = 360.

8 / 20

**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_________?**

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! * 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! * ⁷P₆

9 / 20

**A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most__________?**

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 – 1 = 215.

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 – 1 = 215.

10 / 20

**A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?**

Explanation:

Total number of persons in the committee = 5 + 6 = 11

Number of ways of selecting group of eight persons = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 ways.

Explanation:

Total number of persons in the committee = 5 + 6 = 11

Number of ways of selecting group of eight persons = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 ways.

11 / 20

**What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected?**

Explanation:

The number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women

= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5

= 30 ways.

Explanation:

The number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women

= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5

= 30 ways.

12 / 20

**A committee has 5 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee?**

Explanation:

The number of ways to select two men and three women = ⁵C₂ * ⁶C₃

= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)

= 200

Explanation:

The number of ways to select two men and three women = ⁵C₂ * ⁶C₃

= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)

= 200

13 / 20

**The number of permutations of the letters of the word ‘MESMERISE’ is___________?**

Explanation:

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.

The letter pattern ‘MESMERISE’ consists of 10 letters of which there are 2M’s, 3E’s, 2S’s and 1I and 1R.

Number of arrangements = 9!/(2!)^{2} 3!

Explanation:

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.

The letter pattern ‘MESMERISE’ consists of 10 letters of which there are 2M’s, 3E’s, 2S’s and 1I and 1R.

Number of arrangements = 9!/(2!)^{2} 3!

14 / 20

**The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?**

Explanation:

The word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144.

Explanation:

The word MEADOWS has 7 letters of which 3 are vowels.

-V-V-V-

As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.

Hence the total ways are 24 * 6 = 144.

15 / 20

**Using all the letters of the word “NOKIA”, how many words can be formed, which begin with N and end with A?**

Explanation:

There are five letters in the given word.

Consider 5 blanks ….

The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.

The number of words = 3! = 6.

Explanation:

There are five letters in the given word.

Consider 5 blanks ….

The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.

The number of words = 3! = 6.

16 / 20

**Using all the letters of the word “THURSDAY”, how many different words can be formed?**

Explanation:

Total number of letters = 8

Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.

Explanation:

Total number of letters = 8

Using these letters the number of 8 letters words formed is ⁸P₈ = 8!.

17 / 20

**How many three letter words are formed using the letters of the word TIME?**

Explanation:

The number of letters in the given word is four.

The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

Explanation:

The number of letters in the given word is four.

The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.

18 / 20

**A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?**

Explanation:

The boy can select one trouser in nine ways.

The boy can select one shirt in 12 ways.

The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

Explanation:

The boy can select one trouser in nine ways.

The boy can select one shirt in 12 ways.

The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

19 / 20

**There are 30 people in a group. If all shake hands with one another , how many handshakes are possible?**

20 / 20

**In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?**