A) 3 | B) 5 |

C) 3! | D) 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.

A) 89,972 | B) 90,720 |

C) 72,000 | D) 81,000 |

Explanation:

The given word **HAPPYHOLI** has 9 letters

These 9 letters can e arranged in **9! ways.**

But here in the given word letters **H & P** are repeated twice each

Therefore, Number of ways these 9 letters can be arranged is

$\frac{\mathbf{9}\mathbf{!}}{\mathbf{2}\mathbf{!}\mathbf{}\mathbf{x}\mathbf{}\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{9}\mathbf{}\mathbf{x}\mathbf{}\mathbf{8}\mathbf{}\mathbf{x}\mathbf{}\mathbf{7}\mathbf{}\mathbf{x}\mathbf{}\mathbf{6}\mathbf{}\mathbf{x}\mathbf{}\mathbf{5}\mathbf{}\mathbf{x}\mathbf{}\mathbf{4}\mathbf{}\mathbf{x}\mathbf{}\mathbf{3}}{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{90}\mathbf{,}\mathbf{720}\mathbf{}\mathbf{ways}\mathbf{.}$

A) 60 | B) 120 |

C) 240 | D) 30 |

Explanation:

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = $\mathbf{5}{\mathbf{P}}_{\mathbf{3}}\mathbf{}\mathbf{ways}\mathbf{.}$

$\Rightarrow 5{\mathrm{P}}_{3}=5\mathrm{x}4\mathrm{x}3=\mathbf{60}\mathbf{}\mathbf{words}\mathbf{.}$

A) 4320 | B) 4694 |

C) 4957 | D) 4871 |

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.**

A) 2520 | B) 5040 |

C) 1260 | D) None |

Explanation:

The **7** letters word 'POVERTY' be arranged in $\mathbf{7}{\mathit{P}}_{\mathbf{7}}$ ways = 7! = 5040 ways.

A) 18 | B) 24 |

C) 36 | D) 32 |

Explanation:

Required Number of ways** = **$\mathbf{4}{\mathbf{C}}_{\mathbf{2}}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{3}{\mathbf{C}}_{\mathbf{2}}\mathbf{}\mathbf{\times}\mathbf{}\mathbf{2}{\mathbf{C}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{36}$** = 36**

A) 8640 | B) 720 |

C) 3620 | D) 4512 |

Explanation:

Let the 6 digits of the required 6 digit number be **abcdef**

Then, the number to be odd number the last digit must be odd digit i.e **3, 5, 7 or 9**

The first digit cannot be** ‘0’ **=> possible digits** = 3, 5, 7, 2, 6, 8**

Remaining 4 places can be of **6 x 5 x 4 x 3** ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= **8640 ways.**

A) 1260 | B) 2520 |

C) 5040 | D) 1080 |

Explanation:

The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is

Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2!

But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = **5040 ways.**

A) 380 | B) 420 |

C) 441 | D) 400 |

Explanation:

Given total number of students in the class = 21

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

Therefore, the total number of greeting cards exchanged by the students = **20 x 21 = 420.**