Self Studies
Self Studies
Self Studies
Self Studies

Permutations and Combinations Test 11

Result Self Studies

Permutations and Combinations Test 11
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The number of different 6-digit numbers that can be formed using the three digits 0,1 and 2 is
    Solution
    in the six places of the digits, the first place can be filled only by $$0$$1 or $$2$$ (if $$0$$ comes in first place, that makes it $$5$$ digit number) and the remaining places can be filled by any one of the $$3$$ digits.
    So, first place has only $$2$$ choices and the remaining $$5$$ places have $$3$$ choices each.
    Hence,the number of different $$6$$-digit numbers that can be formed using the three digits $$0,1$$ and $$2$$ = $$2 \times 3^{5}$$
  • Question 2
    1 / -0
    The total number of $$9-$$digit numbers of different digits is
    Solution
    All number are formed by using the digits $$0,1,2,3,4,5,6,7,8,9$$
    The number of arrangements each consisting of $$9$$ digits $$=^{ 10 }{ P }_{ 9 }=10!$$
    These arrangements includes those which start from $$0$$, 
    clearly all such numbers will be $$8$$ digit number.
    the number of such arrangement $$=^{ 9 }{ P }_{ 8 }=9!$$ 
    Therefor number of all $$9$$ digit number $$=10!-9!=10\left( 9! \right) -9!=9\times 9!$$ 
  • Question 3
    1 / -0
    The numbers of three digit number can be formed by using the digits 0, 1, 2, 3, 7, 9 if any digit can be used in any number of time, is

    Solution
    Consider the 3 digit number as
    - - -
    The total number of digits available are $$0,1,2,3,7,9$$.
    Hence the number of ways in which we can fill the first place will be
    $$6-1$$
    $$=5$$ ...(excluding 0).
    Similarly the second place can be filled in $$6$$ ways.
    The third place can be filled in $$ 6$$ ways.
    Hence in total there will be
    $$5\times 6\times 6$$ numbers
    $$=36\times 5$$

    $$=180$$.
  • Question 4
    1 / -0
    The different six digit numbers whose 3 digits are even and 3 digits are odd is

    Solution
    If first digit is even then select other places from remaining five places for two more even digits.
    This can be $$\displaystyle 4\times ^{5}C_{2}\times 5\times 5.$$ 
    Now the remaining three places are to be fit up by odd digit which can be done in $$\displaystyle 5^{3}$$ ways.
    In this way number of six digit number in which three digit are even and $$3$$ are odd in which three digit are even and $$3$$ are odd in which the first digit is even
    $$=\displaystyle 4\times ^{5}C_{2}\times 5\times 5\times 5^{3}=8\times 5^{6}$$    (0 can't occurs at first place)
    Similarly number of six digit numbers which begin with an odd number/digit and which have $$3$$ even, $$3$$ odd digits 
    $$\displaystyle =5\times 5C_{2}\times 5\times 5\times 5^{3}=5\times10\times  5^{5}=10.5^{6}$$ 
    $$\displaystyle \therefore $$ Total six digit number in which $$3$$ odd $$3$$ even are 
    $$\displaystyle =8\times 5^{6}+10\times 5^{6}=18\times 5^{6}=281250$$
  • Question 5
    1 / -0
    A five-digit number divisible by $$3$$ is to be formed using the digits $$0, 1, 2, 3, 4$$ and $$5$$, without repetition. The total number of ways this can be done is
    Solution

    We know that a number is divisible by $$3$$ if the sum of its digits is divisible by $$3$$. Now out of $$0, 1, 2, 3, 4, 5$$ if we take $$1, 2, 3, 4, 5$$ or $$0, 1, 2, 4, 5$$, then the $$5-digit$$ numbers will be divisible by $$3$$.

    Case I:

    Total number of five-digit numbers formed using the digits $$1, 2, 3, 4, 5$$ is $$5! = 120$$.

    Case II:

    Taking $$0, 1, 2, 4, 5$$, total number is $$4\times 4! = 96$$.

    From case I and case II, total number divisible by $$3$$ is $$120 + 96 = 216$$.

  • Question 6
    1 / -0
    If $$m=^{n}\textrm{C}_{2}$$, then $$^{m}\textrm{C}_{2}$$ equals
    Solution
    $$\displaystyle m=^{n}C_{2}=\frac{n\left ( n-1 \right )}{2}...\left ( \ast  \right )$$
    $$\displaystyle

    \therefore ^{m}C_{2}=\frac{m\left ( m-1 \right )}{2}=\frac{\frac{n\left

    ( n-1 \right )}{2}\left [ \frac{n\left ( n-1 \right )}{2}-1 \right

    ]}{2}$$ using (*)
    $$\displaystyle =\frac{n\left ( n-1 \right )\left (

    n^{2}-n-2 \right )}{8}=\frac{n\left ( n-1 \right )\left ( n-2 \right

    )\left ( n+1 \right )}{8}$$
    $$=\dfrac{3\left ( n+1 \right )\left ( n \right

    )\left ( n-1 \right )\left ( n-2 \right )}{4!}$$
    $$ =3.  ^{n+1}C_{4}$$
  • Question 7
    1 / -0
    The number of even numbers with three digits such that if 3 is one of the digit then 5 is the next digit are 
    Solution
    Number of 3 digit even numbers with no $$3=8\times 9\times 5=360$$

    $$8$$ because 1st place can't be 0 and 3 is not included.

    No of 3 digit even number with 3 and hence the constraints$$=1\times 1\times 5=5$$

    note that 2nd and 3rd position can't be 3 since in that case the number can't be even and fulfilling constraint at the same time .

    Therefore, the number of such numbers $$=(360+5)=365$$

    Hence, option 'C' is correct.
  • Question 8
    1 / -0
    The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is 
    Solution
    Letters appearing in the word COCHIN are

    C,C,H,I,N,O

    the words include the words of the form CC---- , CH---- , CI----, CN----  and the first words starting with CO---- is COCHIN 

    thus,the number of words before COCHIN is $$(4)(4!)=96$$

    Hence Option $$C$$ is correct
  • Question 9
    1 / -0
    The number of ways in which one or more letters be selected from the letters $$AAAABBCCCDEF$$ is
    Solution
    The number of ways of selecting
    $$A=4+1=5, B=2+1=3, C=3+1=4,D=1+1=2, E=1+1=2,F=1+1=2$$
    So total ways of selecting is $$5\times 4\times 3\times 2\times 2\times 2=480$$
    Above count also includes the one count when no letter is selected. we need cases when atleast one letter is selected so number of ways$$=480-1=479$$
  • Question 10
    1 / -0
    The domain and range of the function $$\displaystyle f\left ( x \right ) = \sqrt{^{x^{2}+4x}C_{2x^{2}+3}}$$ are
    Solution
    $$\displaystyle f\left ( x \right ) = \sqrt{^{x^{2}+4x}C_{2x^{2}+3}}$$
    For this function to exist $$x^2+4x \geq 0 \Rightarrow  x \in R - [-4,0]$$
    and $$x^2+4x\geq 2x^2+3\Rightarrow x^2-4x+3 \leq 0 \Rightarrow (x-3)(x-1) \leq 0 \Rightarrow x = 1,2,3$$
    Hence domain is $$1,2,3$$ and range is $$\sqrt{^5C_5}, \sqrt{^{13}C_{11}}, \sqrt{^{21}C_{21}} = 1, 2\sqrt{3}$$
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Selfstudy
Selfstudy
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now