SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 1–10 (part 1 of 7)

  1. Exercise 1

    Eight chairs are numbered 1\displaystyle 1 to 8. Two women and 3\displaystyle 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1\displaystyle 1 to 4\displaystyle 4 and then men select from the remaining chairs. Find the total number of possible arrangements. [Hint: 2\displaystyle 2 women occupy the chair, from 1\displaystyle 1 to 4\displaystyle 4 in 4P2\displaystyle { }^4 \mathrm{P}_2 ways and 3\displaystyle 3 men occupy the remaining chairs in 6P3\displaystyle { }^6 \mathrm{P}_3 ways.]

    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
    $\displaystyle 1440$
    The two women take $\displaystyle 2$ of chairs $\displaystyle 1$ to $\displaystyle 4$: \[{}^4P_2 = 4 \times 3 = 12 \] The $\displaystyle 3$ men take $\displaystyle 3$ of the $\displaystyle 6$ chairs left: \[{}^6P_3 = 6 \times 5 \times 4 = 120 \] \[12 \times 120 = 1440 \] Answer: $\displaystyle 1440$

  2. Exercise 2

    If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT ? [Hint: In each case number of words beginning with A, C, H, I is 5\displaystyle 5!]

    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
    $\displaystyle 481$
    Dictionary order of the letters: A, C, H, I, R, T.Words beginning with A, C, H or I, each \(\displaystyle 5!\): \[4 \times 5! = 480 \] Among words beginning with R, RACHIT takes the smallest letter left at every place (A, C, H, I, T), so it comes first. \[\text{rank} = 480 + 1 = 481 \] Answer: $\displaystyle 481$

  3. Exercise 3

    A candidate is required to answer 7\displaystyle 7 questions out of 12\displaystyle 12 questions, which are divided into two groups, each containing 6\displaystyle 6 questions. He is not permitted to attempt more than 5\displaystyle 5 questions from either group. Find the number of different ways of doing questions.

    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
    $\displaystyle 780$
    Splits of the $\displaystyle 7$ questions with at most $\displaystyle 5$ from either group: \(\displaystyle (2,5), (3,4), (4,3), (5,2)\). \[{}^6C_2\,{}^6C_5 + {}^6C_3\,{}^6C_4 + {}^6C_4\,{}^6C_3 + {}^6C_5\,{}^6C_2 \] \[= 15 \cdot 6 + 20 \cdot 15 + 15 \cdot 20 + 6 \cdot 15 \] \[= 90 + 300 + 300 + 90 = 780 \] Answer: $\displaystyle 780$

  4. Exercise 4

    Out of 18\displaystyle 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point. [Hint: Number of straight lines =18C2−5C2+1\displaystyle ={ }^{18} \mathrm{C}_2-{ }^5 \mathrm{C}_2+1.]

    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
    $\displaystyle 144$
    Every pair of points gives a line, except that the $\displaystyle 5$ collinear points give only one line instead of \(\displaystyle {}^5C_2\). \[{}^{18}C_2 - {}^5C_2 + 1 \] \[= 153 - 10 + 1 = 144 \] Answer: $\displaystyle 144$

  5. Exercise 5

    We wish to select 6\displaystyle 6 persons from 8\displaystyle 8, but if the person A is chosen, then B must be chosen. In how many ways can selections be made?

    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
    $\displaystyle 22$
    Case $\displaystyle 1$: A is chosen, so B is chosen; $\displaystyle 4$ more from the other 6. \[{}^6C_4 = 15 \] Case $\displaystyle 2$: A is not chosen; $\displaystyle 6$ from the other 7. \[{}^7C_6 = 7 \] \[15 + 7 = 22 \] Answer: $\displaystyle 22$

  6. Exercise 6

    How many committee of five persons with a chairperson can be selected from 12\displaystyle 12 persons. [Hint: Chairman can be selected in 12\displaystyle 12 ways and remaining in 11C4\displaystyle { }^{11} \mathrm{C}_4.]

    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
    $\displaystyle 3960$
    The chairperson can be any of the $\displaystyle 12$; the other $\displaystyle 4$ members come from the remaining 11. \[12 \times {}^{11}C_4 \] \[= 12 \times 330 = 3960 \] Answer: $\displaystyle 3960$

  7. Exercise 7

    How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

    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
    $\displaystyle 4,68000$
    Two different letters, then three different digits: \[{}^{26}P_2 \times {}^{10}P_3 \] \[= (26 \times 25) \times (10 \times 9 \times 8) \] \[= 650 \times 720 = 468000 \] Answer: $\displaystyle 468000$

  8. Exercise 8

    A bag contains 5\displaystyle 5 black and 6\displaystyle 6 red balls. Determine the number of ways in which 2\displaystyle 2 black and 3\displaystyle 3 red balls can be selected from the lot.

    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
    $\displaystyle 200$
    \[{}^5C_2 = 10, \qquad {}^6C_3 = 20 \] Black and red choices are independent: \[{}^5C_2 \times {}^6C_3 = 10 \times 20 = 200 \] Answer: $\displaystyle 200$

  9. Exercise 9

    Find the number of permutations of n\displaystyle n distinct things taken r\displaystyle r together, in which 3\displaystyle 3 particular things must occur together.

    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
    \(\displaystyle { }^{n-3} \mathrm{C}_{r-3}(r-2)!3!\)
    Pick the other \(\displaystyle r-3\) things, then treat the $\displaystyle 3$ that stay together as one block.\[\text{other things: } {}^{n-3}\mathrm{C}_{r-3} \]\[\text{objects to arrange} = (r-3)+1 = r-2 \;\Rightarrow\; (r-2)! \]\[\text{order inside the block} = 3! \]\[N = {}^{n-3}\mathrm{C}_{r-3}\times (r-2)!\times 3! \]\[N = \frac{(n-3)!}{(r-3)!\,(n-r)!}\,(r-2)!\cdot 6 = 6\,(r-2)\,\frac{(n-3)!}{(n-r)!} \]Answer: \(\displaystyle N = {}^{n-3}\mathrm{C}_{r-3}\,(r-2)!\,3! = 6\,(r-2)\,{}^{n-3}\mathrm{P}_{r-3}\)

  10. Exercise 10

    Find the number of different words that can be formed from the letters of the word 'TRIANGLE' so that no vowels are together.

    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
    $\displaystyle 14400$
    Consonants T, R, N, G, L ($\displaystyle 5$ letters); vowels I, A, E ($\displaystyle 3$ letters). Arrange the consonants, then drop the vowels into separate gaps.\[\text{consonants: } 5! = 120 \]\[\text{gaps} = 5+1 = 6 \]\[\text{vowels in 3 of the 6 gaps: } {}^6\mathrm{P}_3 = 6\cdot5\cdot4 = 120 \]\[N = 5!\times{}^6\mathrm{P}_3 = 120\times120 = 14400 \]Answer: $\displaystyle 14400$