A) 36 | B) 25 |

C) 24 | D) 72 |

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1

b. 3, 1, 1

**Case a.** Number of ways of achieving the first option 2 - 2 - 1

Two toys out of the 5 can be selected in $5{C}_{2}$ ways. Another 2 out of the remaining 3 can be selected in $3{C}_{2}$ ways and the last toy can be selected in $1{C}_{1}$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways $5{C}_{2}*3{C}_{2}$= 15 ways

**Case b**. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in $5{C}_{3}$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is $5{C}_{3}$ = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

A) 1,51,200 ways. | B) 5,04,020 ways |

C) 72,000 ways | D) None of the above |

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in **6P4 ways **

Remaining 7 letters can be arranged in** 7!/3! x 2! ways**

Hence, total number of ways = **6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.**

A) 720 | B) 5040 |

C) 360 | D) 180 |

Explanation:

The number of letters in the given word RITUAL = 6

Then,

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

**=> 6 x 5 x 4 x 3 x 2 x 1 = 720**

A) 64 | B) 63 |

C) 62 | D) 60 |

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

A) 15/52 | B) 17/26 |

C) 13/17 | D) 15/26 |

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

Required, the probability that if a card is drawn either card is black or a king =

$\frac{26}{52}+\frac{4}{52}=\frac{30}{52}=\frac{15}{26}$

A) 48 | B) 24 |

C) 64 | D) 32 |

Explanation:

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2 - 6x5x4/3x2

= 84 - 20

**= 64 ways.**

A) 60 | B) 48 |

C) 36 | D) 20 |

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

**x x x 5**

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

Hence, number 4-digit numbers can be formed are **5 x 4 x 3 = 20 x 3 = 60.**

A) 720 | B) 360 |

C) 120 | D) 24 |

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements** = 4!/2! x 2! = 4! **

**= 4 x 3 x 2 **

**= 24 ways.**

A) 3360 | B) 5040 |

C) 720 | D) 1080 |

Explanation:

Given word is GLACIOUS has 8 letters.

=> **C** is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in **7! ways = 5040.**