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Permutations and Combinations Test 22

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Permutations and Combinations Test 22
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  • Question 1
    1 / -0
    If $$n$$ is an integer greater than $$1$$, then $$a-{ _{  }^{ n }{ C } }_{ 1 }(a-1)+{ _{  }^{ n }{ C } }_{ 2 }(a-2)+....+{ (-1) }^{ n }(a-n)$$ is equal to
    Solution
    $$a-{ _{  }^{ n }{ C } }_{ 1 }(a-1)+{ _{  }^{ n }{ C } }_{ 2 }(a-2)+....+{ (-1) }^{ n }(a-n)$$
    $$=a-{}^{ n }{ C }_{ 1 }a+{}^{n}C_{1}+{}^{ n }{ C }_{ 2 }a-2\ {}^{n}C_{2}+....+(-1)^{n}a-(-1)^{n}n\ {}^{n}C_{n}$$
    $$=a\left\{ { C }_{ 0 }-{ C }_{ 1 }+{ C }_{ 2 }-{ C }_{ 3 }+......{ (-1) }^{ n }{ C }_{ n } \right\} +\left\{ { C }_{ 1 }-{ 2C }_{ 2 }+3{ C }_{ 3 }-......{ (-1) }^{ n }{ n.C }_{ n } \right\}$$
    $$ =a.0+0=0$$
  • Question 2
    1 / -0
    Choose $$3, 4, 5$$ points other than vertices respectively on the sides $$AB, BC$$ and $$CA$$ of a $$\triangle ABC$$. The number of triangles that can be formed by using only these points as vertices, is
    Solution
    Required number of triangles that can be formed by using only given points as vertices
    $$= ^{12}C_{3} - \left \{^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3}\right \}$$
    $$= \dfrac {12\times 11\times 10}{3\times 2\times 1} - \left \{1 + 4 + \dfrac {5\times 4}{2\times 1}\right \}$$
    $$= 220 - (5 + 10)$$
    $$= 220 - 15 = 205$$
  • Question 3
    1 / -0
    Total number of words that can be formed using all letters of the word $$\text{BRIJESH}$$ that neither begins with $$I$$ nor ends with $$B$$ is equal to.
    Solution
    Total number of words formed without any restriction $$=7!$$

    Total number of words beginning with $$I=6!$$

    Total number of words beginning with $$B=6!$$

    Total number of words beginning with I and ends with $$B=5!$$

    Therefore, required number of words $$=7!-6!-6!+5!$$

    $$=5040-2\times 720+120=3720$$
  • Question 4
    1 / -0
    If $$\alpha ,\beta , \gamma $$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also, $$\alpha < \beta < \gamma $$. Then, $$\gamma$$ is equals to:
    Solution
    Let the numbers be $$n-1,n,n+1$$
    As per the given information, we have
    $$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$$ 
    $$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$$
    $$\Rightarrow n=0,1,4$$
    For $$n=0,$$ the numbers are: $$-1,0,1$$  this is out as all the numbers should be positive
    For $$n=1,$$ the numbers are: $$0,1,2$$  this is out as all the numbers should be positive (‘0’ can’t be taken as positive)
    For $$n=4,$$ the numbers are: $$3,4,5$$
    We have got largest number which is $$5$$
    Hence, the value of $$\gamma$$ is $$5$$.
  • Question 5
    1 / -0
    If $$^{n}C_{15} = ^{n}C_{8}$$, then the value of $$^{n}C_{21}$$ is
    Solution
    Given, $${}^{n}C_{15} = {}^{n}C_{8}$$
    $$\Rightarrow {}^{n}C_{n - 15} = {}^{n}C_{8}$$
    $$\therefore n - 15 = 8\Rightarrow n = 23$$
    Now, $${}^{n}C_{21} = {}^{23}C_{21}$$
    $$= \dfrac {(23)!}{2! \times (21)!}$$
    $$= \dfrac {23\times 22}{2} = 23\times 11 = 253$$.
  • Question 6
    1 / -0

    Directions For Questions

    A trip of Saurashtra was organised by the school in a mini bus for 28 students and in a regular bus for 45 students. Each student contributed Rs 500. During the trip, the expenses were : food bill Rs 9,855, entry fees Rs 3,285, mini bus charge Rs 5,432, the regular bus charge Rs 8,730 and boarding charge Rs 6,132. 

    ...view full instructions

    What was the bus charge per student?
    Solution

  • Question 7
    1 / -0
    The value of $$^{50}C_4+\displaystyle\sum^{6}_{r=1}$$ $$^{56-r}C_3$$ is?
    Solution
    We need to find value of $$^{50}C_4+\sum_{r=1}^6\text{ }  ^{56-r}C_3$$
    $$=^{50}C_4+^{50}C_3+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$$
    $$(\because ^nC_i+^nC_{i-1}=^{n+1}C_i)$$
    $$=^{51}C_4+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$$
    $$=^{52}C_4+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$$
    $$=^{53}C_4+^{53}C_3+^{54}C_3+^{55}C_3$$
    $$=^{54}C_4+^{54}C_3+^{55}C_3$$
    $$=^{55}C_4+^{55}C_3$$
    $$=^{56}C_4$$
  • Question 8
    1 / -0
    A village has $$10$$ players. A team of $$6$$ players is to be formed. $$5$$ members are chosen first out of these $$10$$ players and then the captain from the remaining players. Then the total number of ways of choosing such teams is
    Solution
    A village has $$10$$ players.
    Now a team of $$6$$ players is to be formed.
    Thus $$1^{st}$$ we select the $$5$$ players from total $$10$$ players in $${ _{  }^{ 10 }{ C } }_{ 5 }$$ ways and $$6^{th}$$ person will select from the remaining $$5$$ players in $$5$$ ways.
    Total number of ways $$={ _{  }^{ 10 }{ C } }_{ 5 }\times 5=1260$$
  • Question 9
    1 / -0
    An alphabet contains a $$A^{'s}$$ and b $$ B^{'s}$$ . (In all a+b letters ). The number of words each containing all the $$ A^{'s}$$ and any number of $$ B^{'s}$$, is 
    Solution

  • Question 10
    1 / -0
    The maximum number of intersection points of n circles and n straight lines , among themselves  is 80.The value of n is
    Solution

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