PERMUTATION AND COMBINATION 11th class
EXERCISE 3.3
1.Find n, if
$^{n}P_6 : ^{n}P_3=120:1$
Answer :
`\frac{\text{n!}}{\text{(n-6)!}}:\frac{\text{n!}}{\text{(n-4)!}}=120`
`\frac{\text{n!}}{\text{(n-6)!}}:\frac{\text{(n-3)(n-4)(n-5)(n-6)!}}{\text{n!}}=120`
`(n-3)(n-4)(n-5)=120`
`(n-3)(n-4)(n-5)=6 5 4`
so, `n=8`
2.Find m and n, if `^{m+n}P_2=56` and `^{m-n}P_2=12`
Answer:
3. Find r, if `^{12}P_{r-2} : ^{11}P_{r-1}=3:14`
Answer:
`\frac{\text{12!}}{\text{(12-r+2)!}}:\frac{\text{11!}}{\text{(11-r+1)!}}=3:14`
`\frac{\text{12!}}{\text{(12-r+2)!}}:\frac{\text{11!}}{\text{(11-r+1)!}}=3:14`
`\frac{\text{12!}}{\text{(12-r+2)!}}\times\frac{\text{(11-r+1)!}}{\text{11!}}=3:14`
`\frac{\text{12!}}{\text{(14-r)!}}\frac{\text{(12-1)!}}{\text{11!}}=\frac{\text{3}}{\text{14}}`
`\frac{\text{12 11!}}{\text{(14-r)(13-r)(12-r)!}}\frac{\text{(12-r)!}}{\text{11!}}=\frac{\text{3}}{\text{14}}`
`\frac{\text{12}}{\text{(14-r)(13-r)}}=\frac{\text{3}}{\text{14}}`
4. Show that `(n+1) ^nP_r=(n-r+1)\left[^{n+1}P_r\right]`
Answer
`(n+1) ^nP_r=(n-r+1)\left[^{n+1}P_r\right]`
L.H.S=`(n+1)\frac{\text{n!}}{\text{(n-r)!}}`
=`\frac{\text{(n+1)!}}{\text{(n-r)!}}`
R.H.S= `(n-r+1)\frac{\text{(n+1)!}}{\text{(n-r+1)!}}`
= `(n-r+1)\frac{\text{(n+1)!}}{\text{(n-r+1)(n-r)!}}`
`=\frac{\text{(n+1)!}}{\text{(n-r)!}}`
so L.H.S=R.H.S
5.How many 4 letters word can be formed using the letters in the word Madhuri if a letters can be repeated b letters cannot be repeated.
Answer-
6.Determine the number of arrangements of letters of the word algorithm if.
A)vowels are always together.
B)no two vowels are together
C)consonants at even position.
D)O is the first and T is the last letter.
Answer-
A) word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM. When vowels are always together:
There are 3 vowels in the word ALGORITHM. (i.e., A, I, 0)
Let us consider these 3 vowels as one unit. This unit with 6 other letters is to be arranged.
It becomes an arrangement of 7 things which can be done in 7 ways and 3 vowels can be arranged among themselves in `^3P_3` i.e., 3! ways.
Total number of ways in which the word
can be formed=7!×3!=30240
30240 Words can be formed if Vowels are
always together.
B) A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
For no two vowels are together,
There are 6 consonants in the word ALGORITHM.
They can be arranged among themselves in 6P6 i.e., 6! ways.
Let consonants be denoted by C.
_C_C_C_C_C_C_
6 consonants create 7 gaps in which 3
vowels are to arrange.
Therefore, 7 vowels can be filled in `^7 P_3`
=`\frac{\text{7!}}{\text{(7-3)!}}`
=`\frac{\text{7×6×5×4!}}{\text{(4)!}`
=210 ways
Therefore, total number of ways in which the word can be formed
= 6×210
=720×210
=151200
Therefore, 151200 words can be formed if no two vowels are together.
C) A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
When consonants are at even positions,
There are 4 even places and 6 consonants
in the word ALGORITHM.
1st, 2nd, 3rd, 4th even places are filled in 6, 5,4, 3 way respectively.
The number of ways to fill four even
places by consonants 6x 5 x 4 x 3=360
Remaining 5 letters (3 vowels and 2 consonants) can be arranged among themselves in `^5P_5`
i.e., 5! Ways
Therefore,Total number of ways the words can be formed in which even place is occupied by a consonant equal to 360×5!=360×120=43200
Therefore, 43200 words can be formed if even positions are occupied by consonants.
D)A word is to be formed using the letters of the word ALGORITHM.
There are 9 letters in the word ALGORITHM.
When beginning with O and ends with T:
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
7 letters other than O and T can be filled between O and T in 7P7 i.e., 7! ways = 5040 ways.
∴ 5040 words beginning with O and ending with T can be formed.
7.In a group photograph, 6 teachers are in the first row and 18 student are in the second row there are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together find the number of arrangements.
Answer-
In the first row, the middle seat is fixed for the principal.
Also first row, 6 teachers can be arranged among themselves in
In the second row, 12 boys can be arranged among themselves in
13 gaps are created by 12 boys, in which 6 girls are to be arranged. together which can be done in
∴ total number of arrangements
= 6! x 12! is 13P6. using Multiplications
Principle
`= 6! × 12! × \frac{\text{13!}}{(13-6)}`
`=\frac{\text{6!×12!×13!}}{\text{7×6!}}`
`=\frac{\text{12!13!}}{\text{7}}`
8.Find the number of ways so that letter letters of the word history can be arranged as,
a) Y and T are together
b)y is next to T
c)there is no restriction
d) begin and end with vowel
e)end in ST
f)begin with S and end with T
Answer-
a) There are 7 letters in the word HISTORY
When ‘Y’ and ‘T’ are together.
Let us consider ‘Y’ and ‘T’ as one unit
This unit with the other 5 letters is to be arranged.
∴ The number of arrangements of one unit and 5 letters = `6P_6=6`
Also, ‘Y’ and ‘T’ can be arranged among themselves in `^2P_2` i.e., 2! ways.
∴ Total number of arrangements when Y and T are always together = 6! × 2! = 720 × 2 = 1440
∴ 1440 words can be formed if Y and T are together.
b) There are 7 letters in the word HISTORY
When ‘Y’ is next to ‘T’
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 6 letters and one unit = $^6P_6=6$
Also ‘Y’ has to be always next to ‘T’.
So they can be arranged in 1 way.
∴ Total number of arrangements possible when Y is next to T = 6! ×1 = 720
∴ 720 words can be formed if Y is next to T.
c) the word 'HISTORY', the number of letters
is n = 7.
Words in which there is no restriction:
When there is no restriction, the 7 letters of
the word 'HISTORY' can be arranged
among themselves in 'P7 ways.
Hence, the total number of words when
there is no restriction = 7P7 = 7!
d) There are 7 letters in the word HISTORY
When begin and end with vowel
there are 2 vowels in the word HISTORY
All other letters of the word HISTORY are to
be arranged between 2 vowels such that
arrangement begins and ends with vowel.
Other 5 letters can be filled between the two
vowels in 5P5 i.e., 5! Ways
Also, 2 vowels can be arranged among
themselves in 2P2 i.e., 2! Ways
total number of arrangements when the
word begins and ends with vowel
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= 120×2
= 240
Therefore, 240 words can be formed if the arrangement begins and end with vowels.
e) In the word 'HISTORY', the number of letters
is n 7.
Words end in ST:
Since the words end in ST, therefore we
have to arrange only 5 positions from the
remaining 5 letters.
This can be done in 5P5 ways.
Hence, the total number of words end in ST
5P5 = 7
=5!
= 5×4×3×2×1
= 120
f) There are 7 letters in the word HISTORY
When word begins with S and ends with T.
All the letters of the word HISTORY are to be arranged among themselves such that arrangement begins with S and ends with T.
Other 5 letters can be filled between S and T among themselves in 5P` i.e., 5! ways
= 120 ways
∴ 120 words can be formed if arrangement begins with S and ends with T.
9.Find the number of arrangements of the letters in the word SOLAPUR so that consonants and vowels are placed alternately.
Answer:
In the word 'SOLAPUR', the number of
letters is n = 7.
Vowels are 0, A, U, and consonants are S, L,P,R
Since consonants and vowels are placed
alternately, the four consonants must occur
at 4 odd places and three vowels must
occur at 3 even places.
Now, 4 consonants can be arranged in 4P4 ways and
3 vowels can be arranged in 3P3 ways.
Hence, the number of arrangements in which consonants and vowels are placed alternately
4P4 ×3P3
`= 4!3!`
$=4\times3\times2\times1\times3\times2\times1$
=144
10. Find the number of 4 digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
a) digits can be repeated
b) digits cannot be repeated.
Answer-
a) 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be
repeated.
Unit's place digit can be filled in 6 ways.
10's place digit can be filled in 6 ways.
100's place digit can be filled in 6 ways.
1000's place digit can he filled in 6 wavs
There, total number of ways
=6×6×6×6
=`6^{4}`
=1296
b) There are 6 different digits and we have to
form 4-digit numbers, i.e., n = 6, r = 4.
If the repetition of digits is not allowed, then
the number of 4-digit numbers
= `6P4`
=`\frac{\text{ 6!}}{\text{2!}`
=`\frac{\text{ 6×5×4×3×2!}}{\text{2!}`
=360
11.How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that resulting numbers are between 100 and 1000?
Answer-
A number between 100 and 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is of 3 digits and repetition of digits is not allowed.
100's place can be filled in 5 ways as it is a non-zero number which
10's place digits can be filled in 5 ways.
Unit's place digit can be filled in 4 ways.
total number of ways the number can be formed
$= 5\times5\times4$
=100
100 numbers between 100 and 1000 can be formed.
12. Find the number of 6 digit numbers using the digits 3 4 5 6 7 8 without repetition. How many of these numbers are
a) divisible by 5
b) not divisible by 5.
Answer-
There are 6 different digits and we have to form 6-digit numbers, i.e., n = 6, r = 6
If no digit is repeated, the total numbers with 6-digits can be formed =`^6P_6`
=6!
= 6 x 5x4 x3x2x1
=720.
Since the number is divisible by 5, the unit's place of 6-digits number can be filled in only
Since the number is divisible by 5, the unit's place of 6-digits number can be filled in only
one way by the digit 5.
Remaining 5 positions can be filled from the
remaining 5 digits in 5P5 ways.
Hence, the total number of 6-digit numbers
divisible by 5 = 1×5P5 = 5!
=5×4×3×2×1
=120.
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6,7, 8 which
can be done in P6 i.e., 6! = 720 ways
If the number is not divisible by 5
Unit's place can be any digit from 3, 4, 6, 7, 8 which can be selected in 5 ways. Other 5 digits can be arranged in 5P5
i.e., 5! ways
Required number of numbers not by 5
=5×5!
=5×120
=600
13.A code word is formed by two different English letters followed by two non zero distinct digits. Find the number of such codewords. Also, find the number of such code words that end with an even digit.
Answer-
There are 26 alphabets and 9 non-zero
digits from 1 to 9.
For a code, 2 distinct alphabets can be
arranged in 26P2 ways and 2 digits can be arranged in `^9P_2` ways.
Hence, the total number of codewords
available
26P2×9P2
=`\frac{\text{26!}{\text{24!}}`×`\frac{\text{9!}}{\text{2!}}`
=`\frac{\text{26 x 25 x 24! }}{\text{24!}}`×`\frac{\text{9x 8x 7!}}{\text{7!}}`
=(26x 25) x (9 x 8).
=650×72
=46800
For a code, 2 alphabets can be arranged in
26P2 ways, and of the, 2 distinct digits the
last place can be filled by either 2, 4, 6, 8,
i.e., 4 ways, while the first place of the digit
can be filled from the remaining 8 digits in 8
ways.
Hence, the total number of code words that
end with an even integer
$26P2 \times 4\times8$
=26×25×4×8
=20800
14.find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Answer-
There are 5 letters and 3 post boxes and any number of letters can be posted in all of three post boxes.
∴ Each letter can be posted in 3 ways.
∴ Total number of ways 5 letters can be posted
= 3 × 3 × 3 × 3 × 3
= 243
Related Post :- Permutation and Combination Exercise 3.1
15.Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object.
A) always occurs be never occurs.
B) never occurs
Answer-
A) Here n = 11, r = 4
One particular thing always occur:
The number of permutations = rx nPr-1
Putting r = 4, n = 11, we get the number of
permutations
4×`(11-1)P_4`
= 4×`10P_3`
= 4×`\frac{\text{10!}}{\text{7!}}`
= 4x 10 x 9 x8
= 2880
B) There are 11 distinct objects and 4 are to be
taken at a time.
When one particular object will not occur
then 4 object are to be arranged from 10
objects which can be done in \[^10P_4\] ways
=10 x 9 x 8 x 7
=5040
16.In how many ways can 5 different books be arranged on the shelf if
(i)there are no restriction
(ii) 2 books are always together
(iii)2 books are never together
Answer-
(i) When there are no restrictions, the 5 books can be arranged in 5P5 ways.
Hence, the required number of arrangements
= `^5P_5`
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
(ii) 2 books are together
Say books are B1, B2, B3, B4, B5 are to be arranged with B1, B2 together.
Required number = 2x4! = 48
(iii) Since 2 books are never together, we can arrange these two books at 4 places (2 places in between the remaining 3 books +2 at the ends) in P2 ways.
After this the remaining 3 books can be arranged in P3 ways.
Hence, the total number of arrangements in
which 2 books are never together
`^4P_2× ^3P_3`
= 4!×3!
=`\frac{text{4×3×2×1}}{\text{2!}} ×3×2!`
=72.
17. i)3 boys and 3 girls are to sit in a row how many ways can this be done if there are no restriction
ii) there is a girl at each and third boys and girls are at alternate places for all boys sit together.
Answer-
i)As, there are no restriction
Required number=6!=720
ii)The two ends can be occupied by 3 girls in 3P2 ways.
After this the remaining 4 persons can sit at remaining 4 places in 4P4 ways.
Hence, the total number of seating arrangements
= `^3P_2 × ^4P_4`
=3!1!×4!
= 3 × 2 × 1 × 4 × 3 × 2 × 1
= 144.
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| PERMUTATION AND COMBINATION EXERCISE 3.3 11th class MH board |
PERMUTATION AND COMBINATION EXERCISE 3.3 11th class MH board Written By Javeda.
