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Permutations and Combinations Test 42

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Permutations and Combinations Test 42
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  • Question 1
    1 / -0
    A person always prefers to eat parantha and vegetable dish in his meal. How many ways can he make his plate in a marriage party if there are three types of paranthas, four types of vegetable dishes, three types of salads, and two types of sauces?
    Solution
    The number of ways he can select at least one parantha is $$2^3-1=7$$.
    The number of ways he can select at least one vegetable dish is $$2^4-1=15$$.
    The number of ways he can select zero or more items from salads and sauces is $$2^5$$.
    Hence, the total number of ways is $$7\times 15\times 32=3360$$.
  • Question 2
    1 / -0
    Evaluate 
    $$^{47}C_4 + \displaystyle\sum_{j=0}^{3}{^{50-j}C_3} + \sum_{k=0}^{5}{^{56-k}C_{53-k}}$$
    Solution
    Expanding , we get
    $$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}]+[\:^{56}C_{53}+\:^{55}C_{52}+\:^{54}C_{51}+\:^{53}C_{50}+\:^{52}C_{49}+\:^{51}C_{58}]$$
    Now
    $$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$$
    Hence
    $$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}$$
    $$=\:^{48}C_{4}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}$$
    :
    :
    $$=\:^{51}C_{4}$$
    Now
    $$\:^{56}C_{53}+\:^{55}C_{52}+\:^{54}C_{51}+\:^{53}C_{50}+\:^{52}C_{49}+\:^{51}C_{58}$$
    $$=\:^{56}C_{3}+\:^{55}C_{3}+\:^{54}C_{3}+\:^{53}C_{3}+\:^{52}C_{3}+\:^{51}C_{3}$$
    $$\:^{n}C_{r}=\:^{n}C_{n-r}$$
    Hence Adding we get
    $$[\:^{56}C_{3}+\:^{55}C_{3}+\:^{54}C_{3}+\:^{53}C_{3}+\:^{52}C_{3}+\:^{51}C_{3}]+\:^{51}C_{4}$$
    $$=\:^{56}C_{3}+\:^{56}C_{4}$$
    $$=\:^{57}C_{4}$$


  • Question 3
    1 / -0
    For $$2\le r \le n$$, $$\left(\begin{matrix}n \\ r\end{matrix}\right) + 2\left(\begin{matrix}n \\ r-1 \end{matrix}\right) + \left(\begin{matrix}n \\ r - 2 \end{matrix}\right) \space =$$
    Solution
    $$=^nC_r + 2. ^nC_{r-1} + ^nC_{r-2}$$

    $$\displaystyle =^nC_r+\dfrac{2r}{(n-r+1)}.^nC_r+\dfrac{r(r-1)}{(n-r+1)(n-r+2)}.^nC_r$$
    $$\displaystyle =^nC_r\left(\dfrac{n+r+1}{(n-r+1)}+\dfrac{r(r-1)}{(n-r+1)(n-r+2)}\right)$$

    $$\displaystyle =^nC_r\left(\dfrac{n^{2}+3n+2}{(n-r+1)(n-r+2)}\right)$$

    $$\displaystyle =^nC_r\left(\dfrac{(n+1)(n+2)}{(n-r+1)(n-r+2)}\right)$$

    $$=^{n+2}C_r$$

    Hence, option 'D' is correct.
  • Question 4
    1 / -0
    For any positive integer $$m,\space n \space(\mbox{ with } n\ge m) = ^nC_m$$. 
    $$\left(\begin{matrix}n \\ m\end{matrix}\right) + \left(\begin{matrix}n - 1 \\ m\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ m\end{matrix}\right) + ... + \left(\begin{matrix}m \\ m\end{matrix}\right) = $$

    Solution
    $$\begin{pmatrix} m \\ n \end{pmatrix}+\begin{pmatrix} m+1 \\ m \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$$
    $$=\begin{pmatrix} m+1 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+1 \\ m \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$$
    $$=\begin{pmatrix} m+2 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$$
    $$=\begin{pmatrix} m+3 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+3 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$$
    ...................................
    $$=\begin{pmatrix} n \\ m+1 \end{pmatrix}+\begin{pmatrix} n \\ m \end{pmatrix}=\begin{pmatrix} n+1 \\ m+1 \end{pmatrix}$$
    Thus, $$\begin{pmatrix} n \\ m \end{pmatrix}+\begin{pmatrix} n-1 \\ m \end{pmatrix}+\begin{pmatrix} n-2 \\ m \end{pmatrix}+...+\begin{pmatrix} m \\ m \end{pmatrix}=\begin{pmatrix} n+1 \\ m+1 \end{pmatrix}$$
  • Question 5
    1 / -0
    The number of positive integers satisfying the inequality
    $$\quad ^{n+1}C_{n-2} - ^{n+1}C_{n-1} \le 100$$ is
    Solution
    Given, $$\dfrac{(n+1)!}{3!(n-2)!}-\dfrac{(n+1)!}{2!.(n-1)!}$$
    $$=\dfrac{(n-1)(n)(n+1)}{6}-\dfrac{n(n+1)}{2}$$
    $$=(n)(n+1)\dfrac{n-1-3}{6}$$
    $$(n)(n+1)\dfrac{n-4}{6}\leq100$$
    $$\dfrac{n(n+1)(n-4)}{6}\leq 100$$
    Since $$n\epsilon N$$
    Hence
    $$2,3,4,5,6,7,8,9$$ satisfy the above inequality.
  • Question 6
    1 / -0
    The straight lines $$I_1, I_2, I_3$$ are parallel and lie in the same plane. A total number of $$m$$ points on $$I_1$$; $$n$$ points on $$I_2$$; $$k$$ points on $$I_3$$, the maximum number of triangles formed with vertices at these points are
    Solution
    Toatl number of points are $$m+n+k$$
    the $$\Delta 's$$ formed by these points $$=^{ m+n+k }{ { C }_{ 3 } }$$
    Points on the same line gives no triangle
    Such $$\Delta 's$$ are $$^{ m }{ { C }_{ 3 } }+^{ n }{ { C }_{ 3 } }+^{ k }{ { C }_{ 3 } }$$
    Therefore required number $$=^{ m+n+k }{ { C }_{ 3 } }-^{ m }{ { C }_{ 3 } }-^{ n }{ { C }_{ 3 } }-^{ k }{ { C }_{ 3 } }$$
  • Question 7
    1 / -0
    If $$r, s$$ and $$t$$ are prime numbers and $$p, q$$ are positive integers such that the LCM of $$p,q$$ is $$\displaystyle r^{2}t^{4}s^{2}$$ then the number of ordered pair $$(p, q)$$ is 
    Solution
    If 2 numbers A and B have LCM as L, where L = $${ a }^{ x }\times{ b }^{ y }\times{ c }^{ z }$$ ., then, number of ordered pairs of numbers having the above LCM will be: $$(2x + 1) \times (2y + 1) \times (2z + 1)$$

    Let us consider r: we need the power r in either p or q to be at least $$2$$.

    If the power of r in p is 2, then in q it should be 0 or 1 or 2  $$\rightarrow3$$ cases

    If the power of r in q is 2, then in q it should be 0 or 1 or 2 $$\rightarrow3$$ cases

    But (2, 2) has been counted twice. Thus, there are $$3 + 3 - 1 = 5$$ cases

    Total of five cases for the exponents of r such that we have the given LCM which is $$2\times 2 + 1$$

    Similarly, the others follow.
    Here, $$x = 2, y = 4, z = 2.$$

    Therefore, the answer = $$(2\times2+1)\cdot (2\times4+1)\cdot (2\times2+1)=225$$

    Hence, (C) is correct.
  • Question 8
    1 / -0
    If $$\displaystyle ^{7}C_{r} + 3 ^{7}C_{r+1} + 3 ^{7}C_{r+2} + ^{7}C_{r+3} > ^{10}C_{4}$$, then the quadratic equation whose roots are $$\displaystyle \alpha, \: \beta$$ and $$\displaystyle \alpha^{r-1}, \: \beta^{r-1}$$ have
    Solution
    $$\displaystyle ^{7}C_{r}+3^{7}C_{r+1}+3^{7}C_{r+2}+^{7}C_{r+3}> ^{10}C_{4}$$
    $$\displaystyle \Rightarrow\>^{7}C_{r}+\>^{7}C_{r+1}+2\left (\>^{7}C_{r+1}+\>^{7}C_{r+2}  \right )+\>^{7}C_{r+2}+\>^{7}C_{r+3}>\>^{10}C_{4}$$ $$\displaystyle \Rightarrow \>^{8}C_{r+1}+2\>^{8}C_{r+2}+\>^{8}C_{r+3}>\>^{10}C_{4}$$
    $$\displaystyle \Rightarrow \>^{10}C_{r+3}>\>^{10}C_{4}$$
    $$\Rightarrow r+3=5 \Rightarrow r=2$$
    Now roots of two different equation are $$\displaystyle \alpha ,\beta $$ and $$\displaystyle \alpha^{r-1} ,\beta^{r-1}$$
    $$\displaystyle$$ i.e. $$\alpha ^{1},\beta ^{1}$$ and $$\displaystyle \alpha ^{2-1},\beta ^{2-1}$$, i.e. $$\alpha ^{1},\beta ^{1}$$ and $$\displaystyle 
    \alpha ^{1},\beta ^{1}$$ $$\displaystyle \Rightarrow$$  both roots are same
    So Number of common roots are $$ 2$$.
  • Question 9
    1 / -0

    In the figure,two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying, is

    Solution
    The thousands place can be filled by 8 ways in both the cases whereas the other places can be filled by 9 ways in both the cases So, no of ways $$= 2.8.9.9.9 = 11664$$ ways
  • Question 10
    1 / -0
    If $$^nC_{r-1}=36, ^nC_r=84$$ and $$^nC_{r+1}=126$$, then r is
    Solution
    $$^nC_{r-1}=36, ^nC_r=84$$, $$^nC_{r+1}=126$$

    We know that

    $$\dfrac {^nC_{r-1}}{^nC_r}=\dfrac {r}{n-r+1}$$

    or $$\dfrac {36}{84}=\dfrac {r}{n-r+1}$$

    or $$\dfrac {r}{n-r+1}=\dfrac {3}{7}$$

    or $$3n-10r+3=0$$ .....(1)

    Also,

    $$\dfrac {^nC_r}{^nC_{r+1}}=\dfrac {r+1}{n-r}=\dfrac {84}{126}=\dfrac {2}{3}$$

    or $$2n-5r-3=0$$ ....(2)

    Solving (1) and (2), we get $$n=9$$ and $$r=3$$.
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