A) 123 | B) 253 |

C) 321 | D) 231 |

Explanation:

Number of non - negative integral solutions =

=

=

A) 9!/(2!)^{2}x3! | B) 9! x 2! x 3! |

C) 0 | D) None |

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 =

A) ²²C₁₀ + 1 | B) ²²C₉ + ¹⁰C₁ |

C) ²²C₁₀ | D) ²²C₁₀ - 1 |

Explanation:

The total number of ways of forming the group of ten representatives is ²²C₁₀.

The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way

The required number of ways = ²²C₁₀ - 1

A) A | B) B |

C) C | D) D |

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls

or 2 black balls from 4 black balls.

or 3 black balls from 4 black balls.

or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4

= ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls

or 2 red balls from 3 red balls

or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls

= 3C1 + 3C2 + 3C3

= ........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)

or 1 blue ball from 5 blue balls

or 2 blue balls from 5 blue balls

or 3 blue balls from 5 blue balls

or 4 blue balls from 5 blue balls

or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls

= 5C0 + 5C1 + 5C2 + … + 5C5

= ..............(C)

From (A), (B) and (C), required number of ways

= ** **

A) 252 | B) 648 |

C) 243 | D) 900 |

Explanation:

Digits are 0,1,2,3,4,5,6,7,8,9

so no. of digits are 10

first all possible case => 9(0 excluded) x 10 x 10 = 900

second no repetition allowed =>9 x 9 x 8 = 648

third all digits are same => 9 (111,222,333,444,555,666,777,888,999)

Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

A) 133345 | B) 147320 |

C) 13320 | D) 145874 |

Explanation:

Sum of 4 digit numbers = (2+4+6+8) x 3P3 x (1111) = 20 x 6 x 1111 = 133320

Sum of 3 digit numbers = (2+4+6+8) x 3P2 x (111) = 20 x 6 x 111 = 13320

Sum of 2 digit numbers = (2+4+6+8) x 3P1 x (11) = 20 x 3 x 11 = 660

Sum of 1 digit numbers = (2+4+6+8) x 3P0 x (1) = 20 x 1 x 1 = 20

Adding All ,

Sum = 147320