SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Permutations and Combinations

42 questions · 42 still being checked

EXERCISE 6.2 1–5 (part 2 of 5)

  1. Exercise 1

    Evaluate
    (i)
    8\displaystyle 8 ! (ii) 4\displaystyle 4 ! - 3\displaystyle 3 !

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    NCERT’s answer
    (i)
    $\displaystyle 40320$, (ii) $\displaystyle 18$
    Direct expansion. By definition \(\displaystyle n! = n(n-1)(n-2)\cdots 3 \cdot 2 \cdot 1\).(i) \[8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \](ii) \(\displaystyle 4! = 24\) and \(\displaystyle 3! = 6\), so \[4! - 3! = 24 - 6 = 18 \] A neater route is to take out the common factor: \(\displaystyle 4! - 3! = 4 \cdot 3! - 3! = 3!(4-1) = 6 \times 3 = 18\).(i) \(\displaystyle 8! = 40320\) (ii) \(\displaystyle 4! - 3! = 18\).

  2. Exercise 2

    Is 3!+4!=7!\displaystyle 3!+4!=7! ?

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    NCERT’s answer
    $\displaystyle 30$, No
    Evaluate both sides. Never assume the factorial splits over a sum — check by computing.Left side: \[3! + 4! = 6 + 24 = 30 \]Right side: \[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]Since \(\displaystyle 30 \neq 5040\), the two sides are not equal. The tempting step \(\displaystyle 3! + 4! = (3+4)!\) is wrong: the factorial is defined by a product, so it does not distribute over addition.No — \(\displaystyle 3! + 4! = 30\) while \(\displaystyle 7! = 5040\), so \(\displaystyle 3! + 4! \neq 7!\).

  3. Exercise 3

    Compute 8!6!×2!\displaystyle \frac{8!}{6!\times 2!}

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    NCERT’s answer
    $\displaystyle 28$
    Cancel the common factorial. Rather than multiplying \(\displaystyle 8!\) out, write the larger factorial in terms of the smaller one: \[8! = 8 \times 7 \times 6! \]Therefore \[\frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times 2!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]\(\displaystyle \dfrac{8!}{6! \times 2!} = 28\).

  4. Exercise 4

    If 16!+17!=x8!\displaystyle \frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}, find x\displaystyle x

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    NCERT’s answer
    $\displaystyle 64$
    Clear the factorials. Multiply both sides of \[\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!} \] by \(\displaystyle 8!\), the largest factorial present. This turns every term into a whole number. \[\frac{8!}{6!} + \frac{8!}{7!} = x \]Now use \(\displaystyle 8! = 8 \times 7 \times 6!\) and \(\displaystyle 8! = 8 \times 7!\): \[\frac{8!}{6!} = 8 \times 7 = 56, \qquad \frac{8!}{7!} = 8 \]Hence \[x = 56 + 8 = 64 \]\(\displaystyle x = 64\).

  5. Exercise 5

    Evaluate n!(n−r)!\displaystyle \frac{n!}{(n-r)!}, when  (i) n=6,r=2\text { (i) } n=6, r=2 (ii) n=9,r=5\displaystyle n=9, r=5.

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    NCERT’s answer
    (i)
    $\displaystyle 30$, (ii) $\displaystyle 15120$
    Cancel the common factorial. In \(\displaystyle \dfrac{n!}{(n-r)!}\) the denominator is a factor of the numerator, so the quotient is simply the product of the top \(\displaystyle r\) factors of \(\displaystyle n!\).(i) \(\displaystyle n = 6,\ r = 2\). \[\frac{6!}{(6-2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 6 \times 5 = 30 \](ii) \(\displaystyle n = 9,\ r = 5\). \[\frac{9!}{(9-5)!} = \frac{9!}{4!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 = 15120 \](i) $\displaystyle 30$ (ii) 15120.