SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 61–64 (part 7 of 7)

  1. In each if the Exercises from $\displaystyle 60$ to $\displaystyle 64$ match each item given under the column \(\displaystyle \mathrm{C}_1\) to its correct answer given under the column \(\displaystyle \mathrm{C}_2\).

    Exercise 61

    Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
    C1\displaystyle \mathbf{C}_1C2\displaystyle \mathbf{C}_2
    (a) Boys and girls alternate:(i) 5!×6!\displaystyle 5! \times 6!
    (b) No two girls sit together :(ii) 10!−5!6!\displaystyle 10!-5!6!
    (c) All the girls sit together(iii) (5!)2+(5!)2\displaystyle (5!)^2+(5!)^2
    (d) All the girls are never together :(iv) 2!5!5!\displaystyle 2!5!5!

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (a)-(iii), (b)-(i), (c)-(i), (d)-(ii)
    (a)
    Boy first or girl first:
    \[2\times 5!\times 5! = (5!)^2+(5!)^2 = 28800 \]
    (iv)
    \(\displaystyle 2!\,5!\,5!\) is the same number.
    (b)
    Boys in \(\displaystyle 5!\) ways, girls in $\displaystyle 5$ of the $\displaystyle 6$ gaps:
    \[5!\times{}^{6}\mathrm{P}_5 = 5!\times 6! = 86400 \]
    (c)
    Girls as one block, then $\displaystyle 6$ units:
    \[6!\times 5! = 86400 \]
    so (c) matches (i), like (b); \(\displaystyle 2!\,5!\,5!\) would also force the boys into one block.
    (d)
    Complement of (c):
    \[10! - 5!\,6! = 3628800 - 86400 = 3542400 \]
    NCERT prints: (c)-(iv) -- its own (d), \(\displaystyle 10!-5!\,6!\), is the complement of \(\displaystyle 5!\,6!\), not of \(\displaystyle 2!\,5!\,5!\).

  2. Exercise 62

    There are 10\displaystyle 10 professors and 20\displaystyle 20 lecturers out of whom a committee of 2\displaystyle 2 professors and 3\displaystyle 3 lecturer is to be formed. Find :
    C1\displaystyle \mathbf{C}_1C2\displaystyle \mathbf{C}_2
    (a) In how many ways committee : can be formed(i) 10C2×19C3\displaystyle { }^{10} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_3
    (b) In how many ways a particular : professor is included(ii) 10C2×19C2\displaystyle { }^{10} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_2
    (c) In how many ways a particular : lecturer is included(iii) 9C1×20C3\displaystyle { }^9 \mathrm{C}_1 \times{ }^{20} \mathrm{C}_3
    (d) In how many ways a particular : lecturer is excluded(iv) 10C2×20C3\displaystyle { }^{10} \mathrm{C}_2 \times{ }^{20} \mathrm{C}_3

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iv)
    (b)
    \(\displaystyle \leftrightarrow\) (iii)
    (c)
    \(\displaystyle \leftrightarrow\) (ii),
    (d)
    \(\displaystyle \leftrightarrow\) (i)
    (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
    (a)
    Free choice of $\displaystyle 2$ professors and $\displaystyle 3$ lecturers:
    \[{}^{10}\mathrm{C}_2\times{}^{20}\mathrm{C}_3 \]
    (b)
    A particular professor is in; $\displaystyle 1$ more from the other $\displaystyle 9$:
    \[{}^{9}\mathrm{C}_1\times{}^{20}\mathrm{C}_3 \]
    (c)
    A particular lecturer is in; $\displaystyle 2$ more from the other $\displaystyle 19$:
    \[{}^{10}\mathrm{C}_2\times{}^{19}\mathrm{C}_2 \]
    (d)
    A particular lecturer is out; $\displaystyle 3$ from the other $\displaystyle 19$:
    \[{}^{10}\mathrm{C}_2\times{}^{19}\mathrm{C}_3 \]

  3. Exercise 63

    Using the digits 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5, 6\displaystyle 6, 7\displaystyle 7, a number of 4\displaystyle 4 different digits is formed. Find
    C1\displaystyle \mathbf{C}_1C2\displaystyle \mathbf{C}_2
    (a) how many numbers are formed?(i) 840\displaystyle 840
    (b) how many numbers are exactly divisible by 2\displaystyle 2?(ii) 200\displaystyle 200
    (c) how many numbers are exactly divisible by 25\displaystyle 25?(iii) 360\displaystyle 360
    (d) how many of these are exactly divisble by 4\displaystyle 4?(iv) 40\displaystyle 40

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (i)
    (b)
    \(\displaystyle \leftrightarrow\) (iii)
    (c)
    \(\displaystyle \leftrightarrow\) (iv),
    (d)
    \(\displaystyle \leftrightarrow\) (ii)
    (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
    (a)
    $\displaystyle 4$ different digits from $\displaystyle 7$:
    \[{}^{7}\mathrm{P}_4 = 7\cdot 6\cdot 5\cdot 4 = 840 \]
    (b)
    Last digit $\displaystyle 2$, $\displaystyle 4$ or $\displaystyle 6$; the other three places from the remaining $\displaystyle 6$ digits:
    \[3\times{}^{6}\mathrm{P}_3 = 3\times 120 = 360 \]
    (c)
    Ends in $\displaystyle 25$ or $\displaystyle 75$ ($\displaystyle 00$ and $\displaystyle 50$ need a $\displaystyle 0$):
    \[2\times{}^{5}\mathrm{P}_2 = 2\times 20 = 40 \]
    (d)
    Last two digits divisible by $\displaystyle 4$: $\displaystyle 12$, $\displaystyle 16$, $\displaystyle 24$, $\displaystyle 32$, $\displaystyle 36$, $\displaystyle 52$, $\displaystyle 56$, $\displaystyle 64$, $\displaystyle 72$, $\displaystyle 76$ ($\displaystyle 10$ endings):
    \[10\times{}^{5}\mathrm{P}_2 = 10\times 20 = 200 \]

  4. Exercise 64

    How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
    C1\displaystyle \mathbf{C}_1C2\displaystyle \mathbf{C}_2
    (a) 4\displaystyle 4 letters are used at a time(i) 720\displaystyle 720
    (b) All letters are used at a time(ii) 240\displaystyle 240
    (c) All letters are used but the first is a vowel(iii) 360\displaystyle 360

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iii)
    (b)
    \(\displaystyle \leftrightarrow\) (i)
    (c)
    \(\displaystyle \leftrightarrow\) (ii)
    (a)-(iii), (b)-(i), (c)-(ii)
    (a)
    $\displaystyle 4$ of the $\displaystyle 6$ letters, in order:
    \[{}^{6}\mathrm{P}_4 = 6\cdot 5\cdot 4\cdot 3 = 360 \]
    (b)
    All $\displaystyle 6$ letters:
    \[6! = 720 \]
    (c)
    First letter O or A, the other $\displaystyle 5$ letters in any order:
    \[2\times 5! = 240 \]