Test: Permutations & Combinations - Civil Engineering (CE) MCQ

Formula used:

Sum of all the numbers which can be formed by using the n digits without repetition = (n-1)! × (sum of the digits) × (111…..n times)

Where, n → Number of digits

Calculation:

(n-1)! =  (4 - 1)! = 3! = 6

Sum of the digits = 1 + 2 + 3 + 4 = 10

Sum of all the numbers which can be formed by using the n digits without repetition = 6 × 10 × 1111 = 66660

Calculation:
The number of ways of selecting 2 men and 3 women = ⁵C₂ × ⁶C₃

⇒ 5 × 2 × 5 × 4
⇒ 200
∴ The number of ways of selecting 2 men and 3 women from the given committee is 200

First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are
54 + 54 + 54 …
54 (55times) = 54 x 55 = 2970.

The given digits are 1, 3, 5, 7, 9
Sum of r digit number= ^n-1P_r-1
(Sum of all n digits)×(1111… r times)
N is the number of non zero digits.
Here n=5, r=4
The sum of 4 digit numbers
⁴P₃ (1+3+5+7+9)(1111)=666600.

Chess board consists of 9 horizontal 9 vertical lines. A rectangle can be formed by any two horizontal and two vertical lines. Number of rectangles = ⁹C₂ × ⁹C₂ = 1296. For squares there is one 8 by 8 square four 7 by 7 squares, nine 6 by 6 squares and like this
Number of squares on chess board = 1²+2²…..8² = 204
Only rectangles = 1296-204 = 1092.

The number of circular permutations of different things taken all at a time is n-1! and the number of linear permutations of different things taken all at a time is n!.

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are:

33, 35, 37, 53, 55, 57, 73, 75, 77 

Given:
5, 6, 7, 8, 9 are the digits to form 3 digit number

Calculation:
Let us take the 3digit number as H T U (Hundreds, tens, unit digit)  respectively
To make 3 digit number as odd
5, 7, 9 are only possibly be used in the unit digit place
In hundreds and tens place all  5 digits are possible 
Number of ways for unit digit = 3
Number of ways for tens digit = 5
Number of ways for hundreds digit = 5
Number of 3 digits odd number =  3 × 5 × 5 = 75 
∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated

Given 
Total alphabets in word 'FIGHT' = 5 
Concept Used 
Total number ways of arrangement = n! 
Calculation 
The number of different ways of arrangement of n different words (without repetition) = 5! 
⇒ 5 × 4 × 3 × 2 × 1 = 120 
∴ The required answer is 120 

Given:
Word = MANAGEMENT
Calculation:
Vowel occupy 4 places then !4
∵ A and E are repeated then !4/(!2 × !2)
Consonant occupy 6 places then !6
⇒ M and N are repeated then !6/(!2 × !2)

= 6 × 180 = 1080