RD Chapter 16 Permutations Ex 16.5 Solutions
Question - 11 : - How many different numbers, greater than 50000 can be formed with the digits 0, 1, 1, 5, 9.
Answer - 11 : -
Given:
The digits 0, 1, 1, 5,9
Total number of digits= 5
So now, number greaterthan 50000 will have either 5 or 9 in the first place and will consists of 5digits.
Number of 5 digit numbersat first place = 4! / 2! [Since, 1 is repeated]
= [4×3×2!] / 2!
= 4 × 3
= 12
Similarly, number of 9digit numbers at first place = 4! / 2! = 12
The required number ofNumbers = 12 + 12 = 24
Hence, 12 differentnumbers can be formed.
Question - 12 : - How many words can be formed from the letters of the word ‘SERIES’ which start with S and end with S?
Answer - 12 : -
Given:
The word ‘SERIES’
There are 6 letters inthe word ‘SERIES’ out of which 2 are S’s, 2 are E’s and the rest all aredistinct.
Now, Let us fix 5letters at the extreme left and also at the right end. So we are left with 4letters of which 2 are E’s.
These 4 letters can bearranged in n!/ (p! × q! × r!) = 4! / 2! Ways.
Required number ofarrangements is = 4! / 2!
= [4×3×2!] / 2!
= 4 × 3
= 12
Hence, a total numberof arrangements of the letters of the word ‘SERIES’ in such a way that thefirst and last position is always occupied by the letter S is 12.
Question - 13 : - How many permutations of the letters of the word ‘MADHUBANI’ do not begin with M but end with I?
Answer - 13 : -
Given:
The word ‘MADHUBANI’
Total number ofletters = 9
A total number ofarrangements of word MADHUBANI excluding I: Total letters 8. Repeating letterA, repeating twice.
The total number ofarrangements that end with letter I = 8! / 2!
= [8×7×6×5×4×3×2!] /2!
= 8×7×6×5×4×3
= 20160
If the word start with‘M’ and end with ‘I’, there are 7 places for 7 letters.
The total number ofarrangements that start with ‘M’ and end with letter I = 7! / 2!
= [7×6×5×4×3×2!] / 2!
= 7×6×5×4×3
= 2520
The total number ofarrangements that do not start with ‘M’ but end with letter I = The totalnumber of arrangements that end with letter I – The total number ofarrangements that start with ‘M’ and end with letter I
= 20160 – 2520
= 17640
Hence, a total numberof arrangements of word MADHUBANI in such a way that the word is not startingwith M but ends with I is 17640.
Question - 14 : - Find the number of numbers, greater than a million that can be formed with the digit 2, 3, 0, 3, 4, 2, 3.
Answer - 14 : -
Given:
The digits 2, 3, 0, 3,4, 2, 3
Total number of digits= 7
We know, zero cannotbe the first digit of the 7 digit numbers.
Number of 6 digitnumber = n!/ (p! × q! × r!) = 6! / (2! 3!) Ways. [2 is repeated twice and 3 isrepeated 3 times]
The total number ofarrangements = 6! / (2! 3!)
= [6×5×4×3×2×1] /(2×3×2)
= 5×4×3×1
= 60
Now, number of 7 digitnumber = n!/ (p! × q! × r!) = 7! / (2! 3!) Ways
The total number ofarrangements = 7! / (2! 3!)
= [7×6×5×4×3×2×1] /(2×3×2)
= 7×5×4×3×1
= 420
So, total numberswhich is greater than 1 million = 420 – 60 = 360
Hence, total number ofarrangements of 7 digits (2, 3, 0, 3, 4, 2, 3) forming a 7 digit number is 360.