PERMUTATION AND COMBINATION SOLVED PROBLEMS 1

1. 9!

2. 8!

3. 9!/2!

4. 9!/(2!x2!)

9!/(2!x2!)

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

1. 120

2. 180

3. 90

4. 240

180

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

1. 8!

2. 7!

3. 9!

4. 120

9!

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

1. 52

2. 52!

3. 52C2

4. 51

52

We are going to select one unique card from 52

the answer is 52C1 =52

1. 900

2. 480

3. 952

4. 999

952

For a number to be multipe of 5, it should end with 0 or 5

number ending with 0(last number is 0) 9x8x7

number ending with last number 5 is 8x8x7 because 0 cannot take first digit

9x8x7+8x8x7

504+448=952

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