**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:** ^{n}P_{r }= n!/ (n-r)!

Means In the Total set of n numbers p is one type and q, r others then ^{n}P_{r}=n!/p!*q!*r! and ^{n}P_{n}= n!

**Examples: **

**Placement:** like assigned seats, winning a race or 1^{st}, 2^{nd} 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 ^{n}C_{r}

^{n}C_{r }= n!/r!*(n-r)!^{n}C_{o}=1^{n}C_{n}=1^{n}C_{r}=^{n}C_{n}-r^{n}C_{a }=^{n}C_{b }=>a=b=>a+b=n^{n}C_{0 }^{+ n}C_{1 }+^{n}C_{2 }+^{ n}C_{3 }+……..+^{ n}C_{n }=2^{n}

**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:** ^{n}C_{r }= n!/(n-r)!

^{6}C_{3 }= 6!/(6-3)!

^{6}C_{3 }= 6!/3!

^{6}C_{3 }= 120

**Short Trick:** ^{n}C_{r }= 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:**

^{n}P_{r} = n! / (n-r)!

^{9}P_{5} = 9! / (9-5)!

^{9}P_{5} = 9! / 4!

^{9}P_{5} = 15,120

**Short Trick:** ^{n}P_{r} = 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 = ^{5}P_{5} = 5! = 120

**Short Trick: **^{n}P_{n}

**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 × ^{5}P_{4}= 5 × 5! = 5×120 = 600