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Permutation and Combinations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 43 questions.

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Question 11 Mark
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: atmost 3 girls?
Answer
We have to choosen at most 3 girls
In this case the numbers of possibilities are
0 girl and 7 boys
1 girl and 6 boys
2 girls and 5 boys
3 girls and 4 boys
Number of ways to choose 0 girl and 7 boys $=^{4} \mathrm{C}_{0} \times^{9} \mathrm{C}_{7}$
$=\frac{4 !}{0 !(4-0) !} \times \frac{9 !}{7 ! 2 !}=\frac{4 !}{4 !} \times \frac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1}=\frac{72}{2}=36$
Number of ways of choosing 1 girl and 6 boys $=^{4} \mathrm{C}_{1} \times^{9} \mathrm{C}_{6}$
$\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}=\frac{4 \times 3 !}{3 !} \times \frac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2 \times 1}=336$
Number of ways of choosing 2 girls and 5 boys $=^{4} \mathrm{C}_{2} \times^{9} \mathrm{C}_{5}$
$\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}=\frac{4 !}{2 \times 1 \times 2 \times 1} \times \frac{9 \times 7 \times 8 \times 6 \times 5 !}{5 ! 4 !}=756$
therefore ,Number of choosing 3 girls and 4 boys
$=^{4} \mathrm{C}_{3} \times^{9} \mathrm{C}_{4}$
= $\frac{4 !}{3 !(4-3) !} \times \frac{9 !}{4 !(9-4) !}=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}=504$
Total number of ways in which committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632
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Question 21 Mark
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: atleast 3 girls?
Answer
We have to chosen at least 3 girls
There are two possibilities of making committee choosing at least 3 girls
There are 3 girls and 4 boys or there are 4 girls and 3 boys
Choosing three girls we have
$=^{4} \mathrm{C}_{3} \times^{9} \mathrm{C}_{4}$
= $\frac{4 !}{3 !(4-3) !} \times \frac{9 !}{4 !(9-4) !}=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}=504$
Choosing four girls and 3 boys would be done in $^4 \mathrm{C}_{4}$ ways
And choosing 3 boys would be done in $^{9} \mathrm{C}_{3}$therefore,
Total ways $=^{4} \mathrm{C}_{4} \times^{9} \mathrm{C}_{3}$
$=\frac{4 !}{4 !(4-4) !} \times \frac{9 !}{3 !(9-3) !}=\frac{4 !}{4 ! 0 !} \times \frac{9 !}{3 ! 6 !}=\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}=84$
Thus,total numbers of ways of making the committee are
504 + 84 = 588
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Question 31 Mark
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: exactly 3 girls?
Answer
Given,
Total numbers of girls are 4
Out of which 3 are to be chosen
$\therefore$ Number of ways in which choice would be made = $^{4} \mathrm{C}_{3}$
Numbers of boys are 9 out of which 4 are to be chosen which is given by $^{9} \mathrm{C}_{4}$
Total ways of forming the committee with exactly three girls
$=^{4} \mathrm{C}_{3} \times^{9} \mathrm{C}_{4}$
= $\frac{4 !}{3 !(4-3) !} \times \frac{9 !}{4 !(9-4) !}=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}=504$
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Question 41 Mark
If $^n{C_8}{ = ^n}{C_2}$ . find $^n{C_2}$.
Answer
Here $^n{C_8}{ = ^n}{C_2{}}{ \Rightarrow ^n}C_8{ = ^n}{C_{n - 2}}$ $[{\because ^n}{C_r}{ = ^n}{C_{n - r}}]$
$ \Rightarrow 8 = n - 2[{\because ^n}{C_x}{ = ^n}{C_y} \Rightarrow x = y]$
$ \Rightarrow n = 10$ ${\therefore ^n}{C_2}{ = ^{10}}{C_2} = \frac{{10!}}{{2!8!}} = 45$
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Question 51 Mark
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used but first letter is a vowel?
Answer
Here, it is given that total number of letters in MONDAY = 6
No. of vowels in MONDAY =2(O and A)
$\Rightarrow$ No. of permutations in vowel = $_{1}^{2}P$ = 2
Now, remaining places = 5
Remaining letters to be used = 5
$\Rightarrow$ No. of permutations = $\frac{5}{5} \mathrm{P}=\frac{5 !}{(5-5) !}=\frac{5 !}{0 !}$ = 120
Therefore, total number of permutations = 2 $\times$ 120 = 240
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Question 61 Mark
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used at a time.
Answer
We have, total number of letters in MONDAY = 6
Therefore, No. of letters to be used = 6
$\Rightarrow$ No. of permutations = $\frac{6}{6} \mathrm{P}=\frac{6 !}{(6-6) !}=\frac{6 !}{0 !}$ = 720
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Question 71 Mark
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time.
Answer
We know that the number of letters in MONDAY = 6
No. of letters to be used = 4
No. of permutations = $\frac{6}{4} P=\frac{6 !}{(6-4) !}=\frac{6 !}{2 !}$ = 360
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Question 81 Mark
Find r if $^5P_{r} { = ^6}{P_{r - 1}}$
Answer
$^5P_r { = ^6}{P_{r - 1}}$
$\therefore \frac{{5!}}{{(5 - r)!}} = \frac{{6!}}{{(6- r+1)!}}$
$ \Rightarrow \frac{{5!}}{{(5 - r)!}} = \frac{{6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}$
$ \Rightarrow 1 = \frac{6}{{(7 - r)(6 - r)}}$$ \Rightarrow {r^2} - 13r + 42 = 6$
$ \Rightarrow {r^2} - 13r + 36 = 0$$ \Rightarrow {r^2} - 9r - 4r + 36 = 0$
$ \Rightarrow r(r - 9) - 4(r - 9) = 0$$ \Rightarrow (r - 9)(r - 4) = 0$
$ \Rightarrow $ r = 9 or r = 4
Now r = 9 is not possible because r > n.
Thus r = 4
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Question 91 Mark
Find r if $^5{P_r} = {2^6}{P_{r - 1}}$
Answer
Here $^5{P_r} = {2^6}{P_{r - 1}}$
$\frac{{5!}}{{(5 - r)!}} = 2 \cdot \frac{{6!}}{{(7 - r)!}}$
$ \Rightarrow \frac{{5!}}{{(5 - r)!}} = \frac{{2 \times 6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}$
$ \Rightarrow 1 = \frac{{12}}{{(7 - r)(6 - r)}}$$ \Rightarrow {r^2} - 13r + 42 = 12$
$ \Rightarrow {r^2} - 13r + 30 = 0$. $ \Rightarrow {r^2} - 10r - 3r + 30 = 0$
$ \Rightarrow r(r - 10) - 3(r - 10) = 0$
$ \Rightarrow (r - 10)(r - 3) = 0$
$ \Rightarrow $ r = 10 or r = 3
Now r = 10 is not possible because r > n
Thus r = 3
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Question 101 Mark
In, how many ways can the letters of the word PERMUTATIONS be arranged if the there are always 4 letters between P and S?
Answer
We know that, total letters in the word PERMUTATIONS $=12$.
Only repeated letter is T; 2 times
Here, we have $P$ and $S$ are on $1^{\text {st }}$ and $6^{\text {th }}$ places $P$ and $S$ are on $2^{\mathrm{nd}}$ and $7^{\text {th }}$ places
$P$ and $S$ are on $3^{\text {rd }}$ and $8^{\text {th }}$ places $P$ and $S$ are on $4^{\text {th }}$ and $9^{\text {th }}$ places and
$P$ and $S$ are on $5^{\text {th }}$ and $10^{\text {th }}$ places $P$ and $S$ are on $6^{\text {th }}$ and $11^{\text {th }}$ places
$P$ and $S$ are on $7^{\text {th }}$ and $12^{\text {th }}$ Places
Now we see that $P$ and $S$ can be put in $7$ ways and also $P$ and $S$ can interchange their positions.
Therefore, Number of permutations $=2 \times 7=14$
Now the remaining $10$ places can be filled with remaining 10 letters.
$\therefore$ Number of permutations $=\frac{10!}{2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!}=1814400$
Therefore,total number of permutations $=14 \times 1814400=25401600$
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Question 111 Mark
In how many ways can the letters of the word PERMUTATIONS be arranged if the vowels are all together.
Answer
We know that total number of letters in PERMUTATIONS = 12
Only repeated letter is T; 2 times
No. of vowels in PERMUTATIONS = 5(E, U, A, I, O)
Now, we consider all the vowels together as one.
Number of permutations of vowels = $\frac{5}{5}P$ = 120
Now total number of letters = 12 - 5 + 1 = 8
$\Rightarrow$ No. of permutations = $\frac{\frac{s}{8} p}{2 !}=\frac{8 !}{2(8-8) !}=\frac{8 !}{2}$ = 20160
Thus, total number of permutations = 120 $\times $ 20160 = 2419200
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Question 121 Mark
In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S.
Answer
We know that total number of letters in PERMUTATIONS = 12
Only repeated letter is T; 2 times
First and last letter of the word are fixed as P and S respectively.
Number of letters remaining = 12 - 2 = 10
No. of permutations = $\frac{\frac{10 \mathrm{p}}{2 !}}{2 !}=\frac{10 !}{2(10-10) !}=\frac{10 !}{2}$ = 1814400
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Question 131 Mark
Evaluate: $\frac{{n!}}{{(n - r)!}}$ when, n = 9, r = 5
Answer
Substituting,the value of n and r: $\frac{9 !}{(9-5) !}$
$\Rightarrow \frac{9 !}{4 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}=9 \times 8 \times 7 \times 6 \times 5=15120$
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Question 141 Mark
Evaluate: $\frac{{n!}}{{(n - r)!}}$ when, n = 6, r = 2
Answer
Substi the value of n and r: $\frac{6 !}{(6-2) !}$
$\Rightarrow \frac{6 !}{4 !}=\frac{6 \times 5 \times 4 !}{4 !}=6 \times 5=30$
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Question 151 Mark
If $\frac{1}{{6!}} + \frac{1}{{7!}} = \frac{x}{{8!}}$ find x.
Answer
Here
$\frac{1}{{6!}} + \frac{1}{{7!}} = \frac{x}{{8!}}$
$\Rightarrow \frac{1}{{6!}} + \frac{1}{{7 \times 6!}} = \frac{x}{{8 \times 7 \times 6!}}$
$ \Rightarrow \frac{1}{{6!}}\left[ {1 + \frac{1}{7}} \right] = \frac{1}{{6!}}\left[ {\frac{x}{{8 \times 7}}} \right]$
$\Rightarrow(1+\frac{1}{7})=\frac{x}{56}$
$\Rightarrow\frac{8}{7}=\frac{x}{56}$
$\Rightarrow8=\frac{x}{8}$
$\Rightarrow$ x = 64
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Question 161 Mark
Compute $\frac{{8!}}{{6! \times 2!}}$
Answer
$\frac{{8!}}{{6! \times 2!}} = \frac{{8 \times 7 \times 6!}}{{6!(2 \times 1)}} = \frac{{8 \times 7}}{2} = 28$
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Question 171 Mark
Is 3! + 4! = 7!?
Answer
Here $3! + 4! = 3 \times 2 \times 1 + 4 \times 3 \times 2 \times 1 = $$6 + 24 = 30$
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$
$3! + 4! \ne 7!$
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Question 201 Mark
Given 5 flags of different colours, how many different signals can be generated, if each signal requires the use of 2 flags, one below the other?
Answer
Here the upper place of the flag can be fil1ed in 5 ways by using the 5 flags of different colours. Now the lower place of the flag can be filled in 4 ways by using the remaining 4 flags of different colours.
$\therefore $ Total numbers of signals $ = 4 \times 5 = 20$
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Question 211 Mark
A coin is tossed 3 times and the out comes are recorded. How many possible outcomes are there?
Answer
The coin is tossed first time and the number of outcomes is 2. Now the coin is tossed twice and the number of outcomes is 2 again. Also the coin is tossed thrice and the number of outcomes is 2 again.
$\therefore $ Total number of outcomes in tossing a coin 3 times
$ = 2 \times 2 \times 2 = 8$
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Question 221 Mark
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed.
Answer
The unit place can be filled by any one of the digits 1, 2, 3, 4 and 5. So the unit place can be filled in 5 ways. Now four digits left. So the tens place can be filled in 4 ways. The hundreds place can be filled in 3 ways by the remaining 3 digits because repetition of digits is not allowed.
$\therefore $ Total number of 3-digits numbers $ = 3 \times 4 \times 5 = 60$
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Question 231 Mark
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed
Answer
The unit place can be filled by any one of the digits 1, 2, 3, 4 and 5. So the unit place can be filled in 5 ways. Similarly, the tens place and hundreds place can be filled in 5 ways each because the repetition of digits is allowed.
$\therefore$Total number of 3-digits numbers = 5$\times$5$\times$5= 125
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Question 241 Mark
Compute: $\frac{12 !}{(10 !)(2 !)}$
Answer
We have, $\frac{12 !}{(10 !)(2 !)}$= $\frac{12 \times 11 \times(10 !)}{(10 !) \times(2)}$ = 6 $\times$ 11 = 66
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Question 251 Mark
Compute: $\frac{7 !}{5 !}$
Answer
Here, $\frac{7 !}{5 !}$ = $\frac{7 \times 6 \times 5 !}{5 !}$ = 7 $\times$ 6 = 42
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Question 271 Mark
Evaluate: 7 !
Answer
Ans.7 ! = 1 $\times$ 2 $\times$ 3 $\times$ 4 $\times$ 5 $\times$ 6 $\times$ 7 = 5040
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Question 291 Mark
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least 3 girls?
Answer
We know that the team has to consist of at least 3 girls, the team can consist of
(a) 3 girls and 2 boys, or (b) 4 girls and 1 boy.
Note that the team cannot have all 5 girls, because, the group has only 4 girls.
3 girls and 2 boys can be selected in $^{4} \mathrm{C}_{3} \times^{7} \mathrm{C}_{2}$ ways.
4 girls and 1 boy can be selected in $^{4} \mathrm{C}_{4} \times^{7} \mathrm{C}_{1}$ ways.
Thus, the required number of ways = $^{4} \mathrm{C}_{3} \times^{7} \mathrm{C}_{2}+^{4} \mathrm{C}_{4} \times^{7} \mathrm{C}_{1}$ = 84 + 7 = 91
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Question 301 Mark
A group consist of 4 girls and 7 boys. In how many ways, a team of 5 members be selected, if the team has at least one boy and one girl ?
Answer
We know that, at least one boy and one girl are to be there in every team. Thus, the team can consist of
4 boys and 1 girl
This can be done in $ ^{7} C_{4} \times^{4} C_{1} $ ways.
3 boys and 2 girls
This can be done in $ ^{7} C_{3} \times^{4} C_{2} $ ways.
2 boys and 3 girls
This can be done in $ ^{7} C_{2} \times^{4} C_{3} $ ways.
1 boy and 4 girls
This can be done in $ ^{7} C_{1} \times^{4} C_{4} $
Thus,the required number of ways
= $^{7} \mathrm{C}_{1} \times^{4} \mathrm{C}_{4}+^{7} \mathrm{C}_{2} \times^{4} \mathrm{C}_{3}+^{7} \mathrm{C}_{3} \times^{4} \mathrm{C}_{2}+^{7} \mathrm{C}_{4} \times^{4} \mathrm{C}_{1}$
= 140 + 210 + 84 + 7 = 441
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Question 311 Mark
A group consist of $4$ girls and $7$ boys. In how many ways, a team of $5$ members be selected, if the team has no girl?
Answer
Since, the team will not include any girl, thus, only boys are to be selected. $5$ boys out of $7$ boys can be selected in $^7C_5$ ways.
Thus, the required number of ways = $^{7} \mathrm{C}_{5}=\frac{7 !}{5 ! \ 2 !}=\frac{6 \times 7}{2}=21$
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Question 321 Mark
What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these cards are of the same colour.
Answer
Given,there are 26 cards of red colour and $26$ cards of black colour.
Thus, we have to select 4 cards either from black cards or red cards.
$\therefore$ Required number of ways $=\ ^{26}C_4 +\ ^{26}C_4$
$ =2 \times\ ^{26} C_{4} $
$ =2 \times \frac{26 !}{4 ! 22 !} $
$= 29900$
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Question 331 Mark
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these two are red cards and two are black cards.
Answer
Given,there are 26 red cards and 26 black cards.
Thus, the required number of ways = $^{26} \mathrm{C}_{2} \times^{26} \mathrm{C}_{2}$
$=\left(\frac{26 !}{2 ! 24 !}\right)^{2}=(325)^{2}=105625$
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Question 341 Mark
What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these are face cards.
Answer
Given,there are $12$ face cards and we have to select $4$ cards out of these.
$\therefore$ Required number of ways $=\ ^{12}C_4$
$ =\frac{12 !}{4 ! 8 !}  = 495$
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Question 351 Mark
What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these four cards belong to four different suits.
Answer
There are $13$ cards in each suit.
Thus, there are $^{13}C_1$ ways of choosing $1$ card from $13$ cards of diamond,
$^{13}C_1$ ways of choosing $1$ card from $13$ cards of hearts,
$^{13}C_1$ ways of choosing $1$ card from $13$ cards of clubs,
$^{13}C_1$ ways of choosing $1$ card from $13$ cards of spades.
Thus, by multiplication principle, therefore the required number of ways
$ = ^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} $
$ =13 \times 13 \times 13 \times 13 $
$= 13^4$
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Question 361 Mark
What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these four cards are of the same suit.
Answer
We know that,there are four suits: diamond, club, spade, heart and there are $13$ cards of each suit.
Thus, there are $^{13}C_4$ ways of choosing $4$ diamonds.
Similarly, there are $^{13}C_4$ ways of choosing $4$ clubs,
$^{13}C_4$ ways of choosing $4$ spades and $^{13}C_4$ ways of choosing $4$ hearts.
Therefore The required number of ways = $^{13} \mathrm{C}_{4}+^{13} \mathrm{C}_{4}+^{13} \mathrm{C}_{4}+^{13} \mathrm{C}_{4}$
$=4 \times \frac{13 !}{4 ! 9 !}=2860$
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Question 371 Mark
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the words begin with I and end in P?
Answer
We know that, there are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Thus, the required number of arrangements = $\frac{12 !}{3 ! 4 ! 2 !}$ = 1663200
Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters. Therefore, the required number of arrangements = $\frac{10 !}{3 ! 2 ! 4 !}$ = 12600
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Question 381 Mark
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the vowels never occur together.
Answer
We know that, there are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Thus, The required number of arrangements = $\frac{12 !}{3 ! 4 ! 2 !}$ = 1663200
The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
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Question 391 Mark
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do all the vowels always occur together.
Answer
We know that, there are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Thus, The required number of arrangements = $\frac{12 !}{3 ! 4 ! 2 !}$ = 1663200
There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object E, E, E, E, I for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in $\frac{8 !}{3 ! 2 !}$ ways. Corresponding to each of these arrangements, therefore, the 5 vowels E, E, E, E and I can be rearranged in $\frac{5 !}{4 !}$ways. Thus, by multiplication principle, the required number of arrangements = $\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}$ = 16800
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Question 401 Mark
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements, do the words start with P.
Answer
We know that, there are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different.
Thus, The required number of arrangements = $\frac{12 !}{3 ! 4 ! 2 !}$ = 1663200
Let us fix P at the extreme left position, then, the co the arrangements of the remaining 11 letters. Thus, the required number of words starting with P = $\frac{11 !}{3 ! 2 ! 4 !}$ = 138600
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Question 411 Mark
Find the number of different 8-letter arrangement that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.
Answer
If we have to count those permutations in which all vowels are never together, for this, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, those numbers of permutations in which the vowels are always together.
Thus, the required number is = 8 ! – 6 ! $\times$ 3 ! = 6 ! (7 $\times$ 8 – 6)
= 2 $\times$ 6 ! (28 – 3)
= 50 $\times$ 6 ! = 50 $\times$ 720 = 36000
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Question 421 Mark
Find the value of n such that: $\frac{^{n} \mathrm{P}_{4}}{^{n-1} \mathrm{P}_{4}}=\frac{5}{3}, n>4$
Answer
Here, $\frac{^{n} \mathrm{P}_{4}}{^{n-1} \mathrm{P}_{4}}=\frac{5}{3}$
Thus, 3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4)
or 3n = 5 (n – 4) [as (n – 1) (n – 2) (n – 3) $\ne$ 0, n > 4]
or n = 10
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Question 431 Mark
Find the value of $n$ such that: $^nP_5 = 42\ ^nP_{3,} n > 4$
Answer
Given that $^nP_5 = 42\ ^nP_3$
or $n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2)$
Since $n > 4$ so $n(n – 1) (n – 2) \ne 0$
Thus, by dividing both sides by $n(n – 1) (n – 2),$ we obtain
$(n – 3 (n – 4) = 42$
or $n^2 – 7n – 30 = 0$
or $n^2 – 10n + 3n – 30$
or $(n – 10) (n + 3) = 0$
or $n – 10 = 0$ or $n + 3 = 0$
or $n = 10$
or $n = – 3$
As n cannot be negative, therefore $n = 10$
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