Permutation and Combination Methods Shortcut Tricks and Solutions

Permutation and Combination Methods shortcut Tricks and Solutions with Questions:

The Permutation and Combination is the most important topic for preparation any competitive exam. Here we are providing the all kind of tricks related to the Permutation and Combination. You can refer our page to nearby peoples also for latest updates on Math Aptitude Tricks, Problem and Solution. Firstly we understand what is Permutation and Combination??

Permutation and Combination Methods shortcut Tricks and Solutions:

Permutation: The Permutation is called arrangements of digits, numbers, alphabets, colors and letters. In the Permutation orders matters.

Permutation Formula: nPr = n!/ (n-r)!

Means In the Total set of n numbers p is one type and q, r others then nPr=n!/p!*q!*r! and nPn= n!

Examples:

Placement: like assigned seats, winning a race or 1st, 2nd place etc.

Positions: President, Vice President, Secretary etc.

Specific Job: Hand out markers, pass out papers etc.

Combination: Combination is the selection where order doesn’t matters. We can arrange digits, alphabets, numbers and letters etc. like nCr

• nCr = n!/r!*(n-r)!
• nCo=1
• nCn=1
• nCr= nCn-r
• nCa = nCb =>a=b=>a+b=n
• nC0 + nC1 + nC2 + nC3 +……..+ nCn =2n

Difference between Permutation and Combination:

The Basic Difference between permutations is of order. In permutation order matters but in Combination order doesn’t a matter.

Permutations and Combinations Questions with Solutions with Short Tricks:

Question 1: How many three digit numbers can be formed by using the digits in 835621, if there is no repetition??

Solution: nCr = n!/(n-r)!

6C3 = 6!/(6-3)!

6C3 = 6!/3!

6C3 = 120

Short Trick: nCr = n!/(n-r)!

Question 2: Find the number of different words that can be formed from the word ‘ENGINEER’??

Solution:

Total No. of Permutation = n! / p! × q!, where p = of one type , q = ( of another type ).

Total No. of Permutation = 8!/ 3! × 2!

Total No. of Permutation = 3360

Short Trick: n!/p!*q!

Question 3: How many different 5 digit numbers can be formed by using the digits of the number 713628459??

Solution:

nPr = n! / (n-r)!

9P5 = 9! / (9-5)!

9P5 = 9! / 4!

9P5 = 15,120

Short Trick: nPr = n! / (n-r)!

Question 4: How many numbers of five digits can be formed with the digits 1, 3, 5 7 and 9 no digit being repeated??

Solution:

The no. of digits = 5

Required number = 5P5 = 5! = 120

Short Trick: nPn

Question 5: How many numbers of five digits can be formed with the digits 0, 1, 2, 4, 6 and 8?

Solution:

Required no. of numbers = 5 × 5P4= 5 × 5! = 5×120 = 600