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NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 11–20 (part 2 of 7)

  1. Exercise 11

    Find the number of positive integers greater than 6000\displaystyle 6000 and less than 7000\displaystyle 7000 which are divisible by 5\displaystyle 5, provided that no digit is to be repeated.

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    NCERT’s answer
    $\displaystyle 112$
    The number is \(\displaystyle 6\,\_\,\_\,\_\) and divisible by $\displaystyle 5$, so its last digit is $\displaystyle 0$ or 5. The two middle digits come from the $\displaystyle 8$ digits not yet used.\[\text{last digit } 0:\quad {}^8\mathrm{P}_2 = 8\times7 = 56 \]\[\text{last digit } 5:\quad {}^8\mathrm{P}_2 = 8\times7 = 56 \]\[N = 56+56 = 112 \]Answer: $\displaystyle 112$

  2. Exercise 12

    There are 10\displaystyle 10 persons named P1,P2,P3,…P10\displaystyle \mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \ldots \mathrm{P}_{10}. Out of 10\displaystyle 10 persons, 5\displaystyle 5 persons are to be arranged in a line such that in each arrangement P1\displaystyle P_1 must occur whereas P4\displaystyle P_4 and P5\displaystyle \mathrm{P}_5 do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangement =7C4×5!\displaystyle ={ }^7 \mathrm{C}_4 \times 5! ]

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    \(\displaystyle P_1\) is always in and \(\displaystyle P_4, P_5\) are always out, so choose the other $\displaystyle 4$ from the remaining $\displaystyle 7$ persons, then arrange the $\displaystyle 5$ in a line.\[\text{selections} = {}^7\mathrm{C}_4 = 35 \]\[\text{arrangements} = 5! = 120 \]\[N = {}^7\mathrm{C}_4\times 5! = 35\times120 = 4200 \]Answer: $\displaystyle 4200$

  3. Exercise 13

    There are 10\displaystyle 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. [Hint: Required number =210−1\displaystyle =2^{10}-1 ].

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    Each lamp is on or off; the hall is dark only when all $\displaystyle 10$ are off.\[\text{all on/off patterns} = 2^{10} = 1024 \]\[\text{dark hall} = 1 \]\[N = 2^{10}-1 = 1023 \]Answer: $\displaystyle 1023$

  4. Exercise 14

    A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw. [Hint: Required number of ways =3C1×6C2+3C2×6C2+3C3\displaystyle ={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2+{ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2+{ }^3 \mathrm{C}_3.]

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    Total draws minus draws with no black ball (all three from the $\displaystyle 6$ non-black balls).\[\text{all draws} = {}^9\mathrm{C}_3 = 84 \]\[\text{no black} = {}^6\mathrm{C}_3 = 20 \]\[N = 84-20 = 64 \]Check by cases ($\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$ black balls):\[{}^3\mathrm{C}_1\,{}^6\mathrm{C}_2 + {}^3\mathrm{C}_2\,{}^6\mathrm{C}_1 + {}^3\mathrm{C}_3 = 45+18+1 = 64 \]Answer: $\displaystyle 64$

  5. Exercise 15

    If nCr−1=36,nCr=84\displaystyle { }^n \mathrm{C}_{r-1}=36,{ }^n \mathrm{C}_r=84 and nCr+1=126\displaystyle { }^n \mathrm{C}_{r+1}=126, then find rC2\displaystyle { }^r \mathrm{C}_2. [Hint: Form equation using nCrnCr+1\displaystyle \frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r+1}} and nCrnCr−1\displaystyle \frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r-1}} to find the value of r\displaystyle r.]

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    NCERT’s answer
    \(\displaystyle r=3\)
    Use the ratios of consecutive binomial coefficients.\[\frac{{}^n\mathrm{C}_r}{{}^n\mathrm{C}_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac73 \;\Rightarrow\; 3n-10r = -3 \]\[\frac{{}^n\mathrm{C}_{r+1}}{{}^n\mathrm{C}_r} = \frac{n-r}{r+1} = \frac{126}{84} = \frac32 \;\Rightarrow\; 2n-5r = 3 \]Eliminate \(\displaystyle n\):\[6n-20r = -6,\quad 6n-15r = 9 \;\Rightarrow\; 5r = 15 \]\[r = 3,\qquad n = \frac{10r-3}{3} = 9 \]\[{}^9\mathrm{C}_2 = 36,\ {}^9\mathrm{C}_3 = 84,\ {}^9\mathrm{C}_4 = 126 \quad\text{(as given)} \]\[{}^r\mathrm{C}_2 = {}^3\mathrm{C}_2 = 3 \]Answer: $\displaystyle 3$

  6. Exercise 16

    Find the number of integers greater than 7000\displaystyle 7000 that can be formed with the digits 3\displaystyle 3, 5\displaystyle 5, 7\displaystyle 7, 8\displaystyle 8 and 9\displaystyle 9 where no digits are repeated. [Hint: Besides 4\displaystyle 4 digit integers greater than 7000\displaystyle 7000, five digit integers are always greater than 7000.]

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    NCERT’s answer
    $\displaystyle 192$
    Every $\displaystyle 5$-digit number is above $\displaystyle 7000$; a $\displaystyle 4$-digit one must start with $\displaystyle 7$, $\displaystyle 8$ or $\displaystyle 9$; numbers with fewer digits are below 7000.\[\text{5-digit: } 5! = 120 \]\[\text{4-digit: } 3\times{}^4\mathrm{P}_3 = 3\times24 = 72 \]\[N = 120+72 = 192 \]Answer: $\displaystyle 192$

  7. Exercise 17

    If 20\displaystyle 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

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    NCERT’s answer
    $\displaystyle 190$
    Every two lines meet in exactly one point (none parallel), and no point lies on three lines, so each intersection point corresponds to one pair of lines. \[\text{points} = {}^{20}\mathrm{C}_2 \] \[= \frac{20 \times 19}{2} \] \[= 190 \] Answer: $\displaystyle 190$ points

  8. Exercise 18

    In a certain city, all telephone numbers have six digits, the first two digits always being 41\displaystyle 41 or 42\displaystyle 42 or 46\displaystyle 46 or 62\displaystyle 62 or 64\displaystyle 64 . How many telephone numbers have all six digits distinct?

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    NCERT’s answer
    $\displaystyle 8400$
    Each of the $\displaystyle 5$ prefixes has two distinct digits. The last $\displaystyle 4$ places take distinct digits from the other 8. \[\text{prefix choices} = 5 \] \[\text{last four digits} = {}^{8}\mathrm{P}_4 = 8 \times 7 \times 6 \times 5 = 1680 \] \[\text{total} = 5 \times 1680 = 8400 \] Answer: $\displaystyle 8400$

  9. Exercise 19

    In an examination, a student has to answer 4\displaystyle 4 questions out of 5\displaystyle 5 questions; questions 1\displaystyle 1 and 2\displaystyle 2 are however compulsory. Determine the number of ways in which the student can make the choice.

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    NCERT’s answer
    $\displaystyle 3$
    Questions $\displaystyle 1$ and $\displaystyle 2$ are fixed; the other $\displaystyle 2$ answers come from questions $\displaystyle 3$, $\displaystyle 4$, 5. \[{}^{3}\mathrm{C}_2 = 3 \] Answer: $\displaystyle 3$ ways

  10. Exercise 20

    A convex polygon has 44\displaystyle 44 diagonals. Find the number of its sides. [Hint: Polygon of n\displaystyle n sides has (nC2−n)\displaystyle \left({ }^n \mathrm{C}_2-n\right) number of diagonals.]

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    NCERT’s answer
    $\displaystyle 11$
    Diagonals of an \(\displaystyle n\)-sided polygon equal \(\displaystyle {}^{n}\mathrm{C}_2 - n\). \[{}^{n}\mathrm{C}_2 - n = 44 \] \[\frac{n(n-1)}{2} - n = 44 \] \[n^2 - 3n - 88 = 0 \] \[(n - 11)(n + 8) = 0 \] \[n = 11 \quad (n > 0) \] \[\text{check: } {}^{11}\mathrm{C}_2 - 11 = 55 - 11 = 44 \] Answer: $\displaystyle 11$ sides