MATHS 1 Marks Question

If all 11 persons are to be seated in a round table and in the round table of 11 positions there are exactly 5 places for women and 6 places for men.
It is given that no two women sit together, ⁵P₅ this can be done in ways.
The remaining 6 positions can be filled by the 6 men in ⁶P₆ ways. so, by the fundamental principle of counting, the number of seating arrangements as required is ⁵P₅ × ⁶P₆ = 5! × 6!
Hence, the number of ways in which 6 men and 5 worn en can dine at a round table are 5! × 6!.

Given digits are 1, 2, 3, 4

Number of numbers formed by all four digits = ⁴P₄

= 4 × 3 × 2 × 1

Required Numbars = 24

First we will find the least factorial term divisible by 14.
As 7!=7 × 6 × 5! is divisible by 14 leaving remainder zero.
Hence terms 7! onwards can be written as multiple of 7!.
8!=8 × 7!, 9! = 9 × 8 × 7!... like ways 200! can also be written as multiple of 7!.
So all the terms 7! onwards are divisible by 14 leaving remainder zero.
= 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
Hence remainder obtained when 1! + 2! + 3! + ….+ 200! is divided by 14 is 5.

There dice are thrown.
One dice has 6 possible outcomes including 3 even and 3 odd numbers.
Number of outcomes in which at least are dice shows even number
= (one dice shows even numbers) or ( 2 dice shows even number) or (There dice shows even numbers).
= (3 × 6 × 6) + (3 × 3 × 6) + (3 × 3 × 3)
= 108 + 54 + 27
= 189
Required Number = 189

Each of the seven men can be arranged amongst themselves in 7! ways.

The women can be arranged amongst themselves in seven places, in 6! ways (i.e. nothings can be arranged in (n - 1)! ways around a round table). By fundamental principle of counting, total number of ways = 7! × 6!