๐ Permutations & Combinations · Exercises
Example 8 #ch5eg8
Evaluate \(^{10}P_{5} + {}^{10}P_{0}\).
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Example 9 #ch5eg9
Solve the equations for \(n\).
(a) \({}^nP_2 = 9n\) (b) \({}^nP_3 = 12 \cdot {}^nP_2\).
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(a) \begin{align*} {}^nP_2 &= 9n \\ n \ge 2 \quad \text{and} \quad n(n - 1) &= 9n \\ n - 1 &= 9 \\ n &= 10 \end{align*}
(b) \begin{align*} {}^nP_3 &= 12 \cdot {}^nP_2 \\ n \ge 3 \quad \text{and} \quad n(n - 1)(n - 2) &= 12n(n - 1) \\ n - 2 &= 12 \\ n &= 14 \end{align*}
Example 10 #ch5eg10
In how many ways can a president, a treasurer and a secretary for a committee be selected from a group of 15 people?
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The number of ways to select people for three different positions (ranks) is the number of permutations of 15 people taken 3 at a time, and hence it is \[ {}^{15}P_{3} = 15 \cdot 14 \cdot 13 = 2730. \]
Example 11 #ch5eg11
In how many ways can all the letters of the word PENCIL be arranged, without repeating any letters?
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There are 6 distinct letters. So the number of ways to arrange the letters is \[ {}^{6}P_{6} = 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720. \]
Example 12 #ch5eg12
There are 2 buses, which have 5 and 4 vacant seats respectively, and 4 people at a bus stop. In how many ways can all these people be seated on either of the buses, but not both?
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The first bus has 5 vacant seats, so the number of ways to be seated there is \[ {}^{5}P_{4} = 5 \cdot 4 \cdot 3 \cdot 2 = 120. \] The second bus has 4 vacant seats, so the number of ways to be seated there is \[ 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24. \] Therefore, the required number of ways is \(120 + 24 = 144.\)
Example 13 #ch5eg13
In how many ways can 6 different books be arranged along a line on a shelf if one of the books is a dictionary and it must be at one end?
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The dictionary must be fixed at the \(1^{\text{st}}\) or the \(6^{===========\text{th}}\) position in the arrangements.
If the dictionary is in the \(1^{\text{st}}\) position, the number of ways to arrange all the other books in remaining positions is \[ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120. \] If the dictionary is in the \(6^{\text{th}}\) position, the number of ways to arrange all the other books in remaining positions is \[ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120. \] So, the required number of ways \(= (1 \times 120) + (120 \times 1) = 240.\)
Question 1. Short Answer(a) n=7 (b) n=11
Question 2. Short Answer¹⁴P₃ = 14·13·12 = 2184
Question 3. Short Answer⁹P₄ = 9·8·7·6 = 3024
Question 4. Short Answer¹²P₄ = 12·11·10·9 = 11880
Question 5. Short Answer 79,027,200
Example 14 #ch5eg14
In how many ways can a committee of 4 people be selected from a group of 10 people?
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The number of ways is \[{}^{10}C_{4} = \dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210.\]
Example 15 #ch5eg15
Evaluate \({}^{21}C_{1}\), \({}^{21}C_{21}\), \({}^{21}C_{19}\) and \({}^{21}C_{2}\).
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Example 16 #ch5eg16
A music class consists of 5 piano players, 7 guitarists and 4 violinists. A band of 1 piano player, 3 guitarists and 2 violinists must be chosen to play at a school concert. In how many ways can the band be chosen?
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From the music class,
1 piano player can be chosen in \({}^{5}C_{1}\) ways,
3 guitarists can be chosen in \({}^{7}C_{3}\) ways, and
2 violinists can be chosen in \({}^{4}C_{2}\) ways.
So, the number of ways to choose a band is
\[
{}^{5}C_{1} \cdot {}^{7}C_{3} \cdot {}^{4}C_{2} = 5 \cdot \dfrac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} \cdot \dfrac{4 \cdot 3}{2 \cdot 1} = 1050.
\]
Example 17 #ch5eg17
Suppose there are 4 black cars and 7 white cars. If all the cars are distinguishable, in how many ways can 3 cars of the same color be chosen?
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To get the same color, the 3 cars must be all black or all white.
Out of 4 black cars, three can be chosen in \({}^{4}C_{3}\) ways.
Out of 7 white cars, three can be chosen in \({}^{7}C_{3}\) ways.
So, the required number of ways is
\[
{}^{4}C_{3} + {}^{7}C_{3} = \dfrac{4 \cdot 3 \cdot 2}{3 \cdot 2 \cdot 1} + \dfrac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 4 + 35 = 39.
\]
Example 18 #ch5eg18
There are 6 different books. In how many ways can the books be given to 3 children, if the youngest wants to receive 3 books, the elder 1 and the eldest 2 respectively.
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For the youngest child, the books can be chosen in \({}^{6}C_{3} = \dfrac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20\) ways.
For the elder child, the books can be chosen in \({}^{3}C_{1} = 3\) ways.
For the eldest child, the books can be chosen in \({}^{2}C_{2} = 1\) ways.
So, the required number of ways is \(20 \cdot 3 \cdot 1 = 60.\)
Example 19 #ch5eg19
In how many ways can 4 fruits be selected out of 9 fruits, so as always to:
(a) include the largest fruit? (Assume that such a largest fruit exists.)
(b) exclude the smallest fruit? (Assume that such a smallest fruit exists.)
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(a) The largest one is included in every selection. The number of ways only depends on the choice of the other three from the remaining 8 fruits, and hence it is \[ {}^{8}C_{3} = \dfrac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56. \]
(b) Since the smallest one is excluded in every selection, we have to select 4 fruits from the remaining 8 fruits. So the required number ways is \[ {}^{8}C_{4} = \dfrac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70. \]
Question 1. Short Answer¹²C₅·⁷C₄ = 792·35 = 27720 ; ¹²C₄·⁸C₅ = 495·56 = 27720
Question 2. Short Answer¹⁰C₇ = 120 (identical candles)
Question 3. Short Answer 20
Question 4. Short Answer⁹C₂ = 36
Question 5. Short AnswerLines: ⁸C₂ = 28 ; Triangles: ⁸C₃ = 56
Question 6. Short Answer 35
Example 20 #ch5eg20
In how many ways can a permutation of all the letters of the word EXCELLENCE be formed?
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In the word EXCELLENCE, there are 10 letters consisting of four E's, one X, two C's, two L's and one N. So number of ways is \[ \dfrac{10!}{4! \, 1! \, 2! \, 2! \, 1!} = \dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{4! \cdot 1 \cdot (2 \cdot 1) \cdot (2 \cdot 1) \cdot 1} = 10 \cdot 9 \cdot 2 \cdot 7 \cdot 6 \cdot 5 = 37800. \]
Example 21 #ch5eg21
How many permutations are there of the letters of the word PROGRAM, if they do not end in: (a) 2R's? (b) MAP?
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In the word PROGRAM, there are 7 letters consisting of one P, two R's, one O, one G, one A and one M, so the number of permutations of the letters is \[ \dfrac{7!}{2!} = \dfrac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2!}{2!} = 2520. \]
(a) The number of permutations ending in two R's is \(5! = 120\),
since the two R's are fixed at the end.
The number of permutations not ending in two R's \(= 2520 - 120 = 2400.\)
(b) The number of permutations ending in MAP is
\[
\dfrac{4!}{2!} = \dfrac{4 \cdot 3 \cdot 2!}{2!} = 12,
\]
since MAP is fixed at the end, and there are two R's in the remaining 4 letters.
The number of permutations not ending in MAP \(= 2520 - 12 = 2508.\)
Example 22 #ch5eg22
If \(A\) is a set containing 9 distinct elements, how many subsets of \(A\) contain
(a) at most 2 elements?
(b) at least 3 elements?
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(a) The number of combinations containing no element, 1 element and 2 elements are \({}^{9}C_{0}\), \({}^{9}C_{1}\) and \({}^{9}C_{2}\) respectively.
So the number of subsets containing at most 2 elements is
\begin{align*}
{}^{9}C_{0} + {}^{9}C_{1} + {}^{9}C_{2} &= 1 + 9 + \dfrac{9 \cdot 8}{2 \cdot 1} \\
&= 1 + 9 + 36 \\
&= 46.
\end{align*}
(b) The number of all the subsets of \(A\) is \(2^{9} = 512.\)
The number of subsets containing at least 3 elements \(= 512 - 46 = 466.\)
Example 23 #ch5eg23
How many 4-digit even numbers, greater than 4000, can be formed using the digits 1, 2, 3, 4 and 5, without repeating any digit?
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Since the numbers are greater than 4000, the first digit must be 4 or 5.
Case 1. If the first digit is 4 and the last digit is 2,
| First digit | Second digit | Third digit | Last digit |
|---|---|---|---|
| 4 | … | … | 2 |
| 1 way | 3 ways | 2 ways | 1 way |
the numbers can be formed in \(1 \cdot 3 \cdot 2 \cdot 1 = 6\) ways.
Case 2. If the first digit is 5 and the last digit is 2 or 4,
| First digit | Second digit | Third digit | Last digit |
|---|---|---|---|
| 5 | … | … | 2 or 4 |
| 1 way | 3 ways | 2 ways | 2 ways |
the numbers can be formed in \(1 \cdot 3 \cdot 2 \cdot 2 = 12\) ways.
So, the number of 4-digit even numbers greater than 4000 is \(6 + 12 = 18.\)
Example 24 #ch5eg24
In how many ways can 2 different chemistry books, 4 different mathematics books and 3 different physics books be arranged in a line on a shelf if
(a) 2 chemistry books are to be placed on the left, 4 mathematics books in the middle and 3 physics books on the right?
(b) books of the same subjects are together?
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(a) The number of ways to place 2 chemistry books in the left part is \(2! = 2 \cdot 1 = 2.\)
The number of ways to place 3 physics books in the right part is \(3! = 3 \cdot 2 \cdot 1 = 6.\)
The number of ways to place 4 mathematics books in the middle part is \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24.\)
The required number of ways \(= 2 \cdot 24 \cdot 6 = 288.\)
(b) The 3 subject wise groups can be placed in \(3! = 6\) ways, and in each such arrangement the books can be placed in 288 ways (as in part (a)).
The required number of ways to arrange books of the same subject together \(= 6 \cdot 288 = 1728.\)
Example 25 #ch5eg25
How many permutations of the letters S, U, N, D, A, Y are there if the two vowels are placed together?
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If the two vowels U and A unite to form a new single letter, then the number of permutations of the letters is \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120.\) Again, U and A can be arranged in 2 different ways as UA and AU in each such permutation.
So the required number of permutations is \(120 \cdot 2 = 240.\)
Example 26 #ch5eg26
Find the number of permutations of all the letters of the word INTERNET in such a way that there are exactly 4 letters between the two T's.
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The letters must be arranged in one of the three forms:
in which the 6 letters (2 N's, 2 E's, 1 I and 1 R) other than 2 T's are to be put in star positions. For each such form, the number of ways to put the six letters in star positions is \[ \dfrac{6!}{2! \cdot 2!} = \dfrac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} = 180. \] So the required number of permutations is \(3 \cdot 180 = 540.\)
Question 1. Short Answer(i) 3360 ; (ii) 30240 ; (iii) 453600
Question 2. Short Answer848,800
Question 3. Short Answer2¹⁰ − 1 = 1023
Question 4. Short Answer 2784
Question 5. Short Answer(i) 4⁴ = 256 (ii) 4! = 24 (iii) 12
(i) there is no restriction?
(ii) repetition is not allowed?
(iii) repetition is not allowed, and 0 is either the first or the last digit?
Question 6. Short Answer(i)30 (ii) 180 (iii) 120 (iv) 120
(i) starting with T and ending with N?
(ii) starting and ending with a consonant?
(iii) with vowels only?
(iv) if it contains 3 consonants?
Question 7. Short Answer(i) 144 (ii) 72
(i) with 3 sisters standing together?
(ii) if brothers and sisters are in alternating positions?
Question 8. Short Answer(i) 720 (ii) 4320 (iii) 4896
(i) the 3 vowels are placed together?
(ii) the 3 vowels are not placed together?
(iii) consonants and vowels do not appear alternately?
Question 9. Short Answer 360
Question 10. Short Answer (i) 1344 (ii) 13440.
(i) 2 E's.
(ii) 2 S's.
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