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Permutations and Combinations Test 0

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Permutations and Combinations Test 0
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  • Question 1
    1 / -0
    The no .of ways of selecting $$3$$ men and $$2$$ women from $$6$$ men and $$6$$ women.
    Solution
    The no.o  f ways of selecting Men is $$^6C_3$$
    The no.o  f ways of selecting Women is $$^6C_2$$
    Total no.  of ways is $$^6C_3\times ^6C_2$$
  • Question 2
    1 / -0
    How many six letter words be made out of the letters of $$ASSIST$$ ? In how many words the alphabets $$S$$ alternates with other letters ?
    Solution
    The word ASSIST has 3 's' and 3 other characters (AIT).
    Place the other 3 characters first. This can be done in 3! ways.
    Now there are 4 spaces remaining (1234) mark as 1,2,3,4.
    3 's' 5 can be placed at the spaces marked as 1,2,3 or at the spaces marked as 2,3,4 (2)was 
    Total no.of ways =3! .2$$=3\times 2\times 2=12$$
    Total no.of ways =5!.$$=5\times 4\times 3\times 2=120.$$
    $$\therefore$$ so, the answer is 120,12.

  • Question 3
    1 / -0
    The number of ways in which 6 rings can be worn on the four fingers of one hand is 
    Solution
    The four fingers can be selected in $$^5C_4=5$$ ways
    And 6 rings can be placed in $$^6C_4$$ ways
    So total no.of ways are $$5\times ^6C_4$$
  • Question 4
    1 / -0
    If the coefficients of three consecutive terms in the expansion of  $$(1+x)^n$$ are in the ratio of $$1:7:42$$, then $$n$$ is divisible by-
    Solution
    Let the three consecutive terms be $$(r-1)th, rth , (r+1)th $$ terms.
    That is, $$T_{r-1}, T_{r}. T_{r+1}$$.
    We know that general term of expansion $$(a+b)^{n}$$ is
    $$T_{r+1}={n_{C}}_{r}a^{n-r}b^{r}$$
    For $$(1+x)^{n}$$,
    Putting $$a=1, b=x$$,
    $$T_{r+1}={n_{C}}_{r}1^{n-r}x^r$$
    $$T_{r+1}={n_{C}}_{r}x^r$$
    Therefore, coefficient of $$(r+1)th$$ term = $${n_{C}}_{r}$$
    For $$rth$$ term of $$(1+x)^{n}$$
    Replacing r with r-1, we get,
    $$T_{r}={n_{C}}_{r-1}x^{r-1}$$
    Therefore, coefficient of $$rth$$ term = $${n_{C}}_{r-1}$$
    For $$(r-1)th$$ term of $$(1+x)^{n}$$
    Replacing r with r-2. we get,
    $$T_{r-1}={n_{C}}_{r-2}x^{r-2}$$
    Therefore, coefficient of $$(r-1)th$$ term = $${n_{C}}_{r-2}$$
    Since the coefficient of (r-1)th, rth and (r+1)th terms are in the ratio 1 : 7 : 42,
    $$\dfrac{{n_{C}}_{r-2}}{{n_{C}}_{r-1}}=\dfrac{1}{7}$$

    $$\dfrac{\dfrac{n!}{(r-2)![n-(r-2)]!}}{\dfrac{n!}{(r-1)![n-(r-1)]!}}=\dfrac{1}{7}$$

    $$\dfrac{(r-1)[n-(r-1)]!}{[n-(r-2)]!}=\dfrac{1}{7}$$

    $$\dfrac{(r-1)(n-r+1)!}{(n-r+2)!}=\dfrac{1}{7}$$

    $$\dfrac{(r-1)(n-r+1)!}{(n-r+2)(n-r+1)!}=\dfrac{1}{7}$$

    $$\dfrac{(r-1)}{(n-r+2)}=\dfrac{1}{7}$$

    $$n-8r+9=0$$         ......(1)

    Also,
    $$\dfrac{\dfrac{n!}{(r-1)![n-(r-1)]!}}{\dfrac{n!}{r!(n-r)!}}=\dfrac{7}{42}$$

    $$\dfrac{n!}{(r-1)!(n-r+1)!} \times \dfrac{r!(n-r)!}{n!}=\dfrac{1}{6}$$

    $$\dfrac{r(n-r)!}{(n-r+1)!}=\dfrac{1}{6}$$

    $$\dfrac{r}{n+1-r}=\dfrac{1}{6}$$

    $$n-7r+1=0$$      .....(2)

    Solving equations 1 & 2, we get,
    $$r=8$$ and $$n=55$$
    Hence, $$n=55$$
  • Question 5
    1 / -0
    If P (n, n) denotes the number of permutations of n different things taken all at a time then P (n, n ) is also identical to:
    Solution

  • Question 6
    1 / -0
    When we realize a specific implementation of a pancake algorithm, every move when we find the greatest of the sized array and flipping can be modeled through ____________.
    Solution
    Unlike a traditional sorting algorithm, which attempts to sort with the fewest comparisons possible, the goal is to sort the sequence in as few reversals as possible.
    Here when we flipping the array or stack, we have to take utmost priority to preserve the order of the list so that that sorting doesnt become invalid. Hence we use permutations, we are ensuring that order matters.
  • Question 7
    1 / -0
    There are 8 types of pant pieces and $$9$$ types of shirt pieces with a man. The number of ways in which a pair ($$1$$ pant, $$1$$ shirt) can be stitched by the tailor is
    Solution
    Each piece of pant can be paired with $$9$$ different pieces of shirt.
    Now there are $$8$$ pieces of pant.
    Taking one piece of a pant at time, there are $$9$$ ways to stitch it, since there are $$9$$ different pieces of shirt.
    Hence, the total ways will be
    $$=8\times 9$$
    $$=72$$
  • Question 8
    1 / -0
    If $$ ^nP_r = ^nP{_r}{_+}{_1} $$ and  $$ ^nC_r = ^nC{_r}{_-}{_1} $$, then the values of n and r are:
    Solution
    $$^nP_r =  ^nP_{r+1}$$
    $$\dfrac{n!}{(n-r)!} = \dfrac{n!}{(n-(r+1))!}$$
    $$(n-r-1)! = (n-r)!$$
    $$n-r = 1$$      -------------(1)
    $$^nC_r =  ^nC_{r-1}$$
    $$\dfrac{n!}{(n-r)! r!} = \dfrac{n!}{(n-r+1)!(r+1)!}$$
    $$(r-1)!(n+1-r)! = (n-r)!r!$$
    $$n+1-r = r$$      
    $$n+1 = 2r$$
    $$n = 2r-1$$     -------------(2)
    By solving both equations, we get
    $$r = 2$$
    $$\Rightarrow n = 2(2)-1 = 3$$
  • Question 9
    1 / -0
    There are 'mn' letters and n post boxes. The number of ways in which these letters can be posted is:
    Solution
    Every letter can be posted in any of the n post boxes 
    $$\therefore$$Total number of ways $$=n\times n\times n\times........(mn \ times) = n^{mn}$$
  • Question 10
    1 / -0
    The number of words that can be formed using any number of letters of the word "KANPUR" without repeating any letter is
    Solution
    $$(a) 1$$ Letter word:
    Selecting $$1$$ letter of $$6$$ letter $$K,A,N,P,U,R=\quad^{6}{C}_{1}=6$$ ways

    $$(b) 2$$ Letter words:
    Selecting $$2$$ letter of $$6$$ letter $$K,A,N,P,U,R= \quad ^{6}{C}_{2}$$
    Therefore no. of ways to create $$2$$ letter words $$=\quad ^{6}{C}_{2} \times 2!$$

    $$(c) 3$$ Letter words:
    Selecting $$3$$ letter of $$6$$ letter $$K,A,N,P,U,R= \quad ^{6}{C}_{3}$$
    Therefore no. of ways to create $$3$$ letter words $$=\quad ^{6}{C}_{3} \times 3!$$

    $$(d) 4$$ Letter words:
    Selecting $$4$$ letter of $$6$$ letter $$K,A,N,P,U,R= \quad ^{6}{C}_{4}$$
    Therefore no. of ways to create $$4$$ letter words $$=\quad ^{6}{C}_{4} \times 4!$$

    $$(e) 5$$ Letter words:
    Selecting $$5$$ letter of $$6$$ letter $$K,A,N,P,U,R= \quad ^{6}{C}_{5}$$
    Therefore no. of ways to create $$5$$ letter words $$=\quad ^{6}{C}_{5} \times 5!$$

    $$(f) 6$$ Letter words:
    Selecting $$6$$ letter of $$6$$ letter $$K,A,N,P,U,R= \quad ^{6}{C}_{6}$$
    Therefore no. of ways to create $$6$$ letter words $$=\quad ^{6}{C}_{6} \times 6!$$

    Total $$= 6+\quad^{6}{C}_{2}\times 2!+\quad ^{6}{C}_{3}\times 3!+\quad ^{6}{C}_{4}\times 4!+\quad ^{6}{C}_{5}\times 5!+ 6!= 1956$$ words
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