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    Subject

    Quantitative Aptitude

    Topic

    Permutation & Combination

    Find the value of6P2×5P3

    ক)
    1600
    খ)
    1860
    গ)
    1470
    ঘ)
    1720
    ঙ)
    None of these

    Explanation

    We know that: Formula: nPr=!n!(n−r){}_{}^n{P_r} = \frac{{!n}}{{!\left( {n - r} \right)}} ∴ 6 P 2 × 5 P 3 =  !6!(6−2)×!5!(5−3)\frac{{!6}}{{!\left( {6 - 2} \right)}} \times \frac{{!5}}{{!\left( {5 - 3} \right)}} ⇒ 6 P 2 × 5 P 3 =!6!4×!5!2= \frac{{!6}}{{!4}} \times \frac{{!5}}{{!2}} ⇒ 6 P 2 × 5 P 3 = 6×5×4×3×2×14×3×2×1×5×4×3×2×12\frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \frac{{5 \times 4 \times 3 \times 2 \times 1}}{2} ⇒ 6 P 2 × 5 P 3 = 30 × 60 ⇒ 6 P 2 × 5 P 3 = 1800

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