# Permutation and Combination

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Question 1 |

What is the number of possible words that can be made using the word “EASYQUIZ” such that the vowels always come together?

120 | |

720 | |

2880 | |

4320 |

**Permutation and Combination**

**Discuss it**

Question 1 Explanation:

The word “EASYQUIZ” has 8 letters in which “EAUI” are vowels.
Since vowels always come together, we can assume “EAUI” as a single unit letter.
4+1 letter can be arranged in 5! ways. Also, vowels “EAUI” can be arranged in 4! ways.
Hence the total number of possible words = 5! * 4! = 2880.

Question 2 |

What is the number of possible words that can be made using the word “QUIZ” such that the vowels never come together?

8 | |

12 | |

16 | |

24 |

**Permutation and Combination**

**Discuss it**

Question 2 Explanation:

The word “QUIZ” has 4 letters in which “UI” are vowels.
Total number of possible words = 4!
Treating “UI” as a single letter we can make words in 3! ways.
“UI” can be arranged in 2! ways.
Therefore, the number of words can be made using vowels together = 3! * 2! = 12 ways.
Hence, the number of words can be made such that vowels never come together = 24 – 12 = 12 ways.

Question 3 |

How many words can be made from the word “APPLE” using all the alphabets with repetition and without repetition respectively?

1024, 60 | |

60, 1024 | |

1024, 1024 | |

240, 1024 |

**Permutation and Combination**

**Discuss it**

Question 3 Explanation:

The word “APPLE” has 5 letters in which “P” comes twice.
If repetition is allowed, the number of words we can form = 4*4*4*4*4 = 1024.
(This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.)
If repetition is not allowed, the number of words we can form = 5!/2! = 60. (This is because "P" comes twice.)

Question 4 |

How many ways a 6 member team can be formed having 3 men and 3 ladies from a group of 6 men and 7 ladies?

700 | |

720 | |

120 | |

500 |

**Permutation and Combination**

**Discuss it**

Question 4 Explanation:

We have to pick 3 men from 6 available men and 3 ladies from 7 available ladies.
Required number of ways =

^{6}C_{3}*^{7}C_{3}= 20 * 35 = 700.Question 5 |

In how many ways can an interview panel of 3 members be formed from 3 engineers, 2 psychologists and 3 managers if at least 1 engineer must be included?

30 | |

15 | |

46 | |

45 |

**Permutation and Combination**

**Discuss it**

Question 5 Explanation:

The interview panel of 3 members can be formed in 3 ways by selecting 1 engineer and 2 other professionals, 2 engineers and 1 other professionals and all 3 engineers.

- 1 engineer out of 3 engineers and 2 other professionals out of 5 professionals can be selected as
=
^{3}C_{1}*^{5}C_{2}= 3 * 10 = 30 ways. - 2 engineers out of 3 engineers and 1 other professional out of 5 professionals can be selected as
=
^{3}C_{2}*^{5}C_{1}= 3 * 5 = 15 ways. - 3 engineers out of 3 engineers and 0 other professional out of 5 professionals can be selected as
=
^{3}C_{3}*^{5}C_{0}= 1 way.

Question 6 |

How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 which are divisible by 5 when none of the digits are repeated?

120 | |

35 | |

24 | |

720 |

**Permutation and Combination**

**Discuss it**

Question 6 Explanation:

A number is divisible by 5 if and only if its last digit is either 5 or 0. But, 0 is not available here. So, we have to fix 5 as a last digit of 4-digit number and fill 3 places with remaining 6 digits.
Number of ways to choose 3 digits =

^{6}C_{3}= 20. Number of ways to arrange the chosen digits = 3! Hence, total number of required ways =^{6}C_{3}* 3! =^{6}P_{3}= 120.Question 7 |

In how many ways can 20 boys and 18 girls make a queue such that no two girls are together?

20!* ^{20}C_{18} | |

20!* ^{20}P_{18} | |

20!* ^{21}C_{18} | |

20!* ^{21}P_{18} |

**Permutation and Combination**

**Discuss it**

Question 7 Explanation:

The boys will be arranged in 20! ways. Now, there are a total of 21 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls).
Therefore, the 18 girls can stand at these 21 places only.
Hence, the number of ways = 20!*

^{21}P_{18}Option (D) is correct.Question 8 |

There are 5 floating stones on a river. A man wants to cross the river. He can move either 1 or 2 steps at a time. Find the number of ways in which he can cross the river?

11 | |

12 | |

13 | |

14 |

**Permutation and Combination**

**Discuss it**

Question 8 Explanation:

The man needs to take 6 steps to cross the river. He can do this in the following ways:

- Crossing the river by 6 unit steps = 1 way.
- Crossing the river by 4 unit steps and 1 double step =
^{5}C_{1}= 5C4 = 5 ways. - Crossing the river by 2 unit steps and 2 double steps =
^{4}C_{2}= 6 ways. - Crossing the river by 3 double steps = 1 way.

Question 9 |

Out of 7 boys and 4 girls, how many queues of 3 boys and 2 girls can be formed?

120 | |

25200 | |

24800 | |

1440 |

**Permutation and Combination**

**Discuss it**

Question 9 Explanation:

Number of ways to choose 3 boys out of 7 =

^{7}C_{3}. Number of ways to choose 2 girls out of 4 =^{4}C_{2}. Therefore, number of ways to choose the required groups =^{7}C_{3}*^{4}C_{2}= 35 * 6 = 210. Number of ways to arrange the 3 boys and 2 girls in a queue = 5! = 120. Therefore, the required number of queues = 210 * 120 = 25200.Question 10 |

A box contains 2 red coins, 3 green coins and 4 blue coins. In how many ways can 3 coins be chosen such that at least one coin is green?

16 | |

32 | |

64 | |

128 |

**Permutation and Combination**

**Discuss it**

Question 10 Explanation:

There are three cases:

- 3 green coins
- 2 green coins + 1 non-green coin
- 1 green coin + 2 non-green coins

^{3}C_{3}+^{3}C_{2}*^{6}C_{1}+^{3}C_{1}*^{6}C_{2}= 1 + 3*6 + 3*15 = 64.
There are 16 questions to complete.