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Question -

Convert the following products into factorials:
(i) 5 тЛЕ 6 тЛЕ 7 тЛЕ 8 тЛЕ 9 тЛЕ 10
(ii) 3 тЛЕ 6 тЛЕ 9 тЛЕ 12 тЛЕ 15 тЛЕ 18
(iii) (n + 1) (n + 2) (n + 3) тАж(2n)
(iv) 1 тЛЕ 3 тЛЕ 5 тЛЕ 7 тЛЕ 9 тАж (2n тАУ 1)



Answer -

(i)┬а5┬атЛЕ┬а6┬атЛЕ┬а7┬атЛЕ┬а8┬атЛЕ┬а9┬атЛЕ┬а10

Let us evaluate

We can write it as:

5┬атЛЕ┬а6┬атЛЕ┬а7┬атЛЕ┬а8┬атЛЕ┬а9┬атЛЕ┬а10 = (1├Ч2├Ч3├Ч4├Ч5├Ч6├Ч7├Ч8├Ч9├Ч10)/(1├Ч2├Ч3├Ч4)

= 10!/4!

(ii)┬а3┬атЛЕ┬а6┬атЛЕ┬а9┬атЛЕ┬а12┬атЛЕ┬а15┬атЛЕ┬а18

Let us evaluate

3┬атЛЕ┬а6┬атЛЕ┬а9┬атЛЕ┬а12┬атЛЕ┬а15┬атЛЕ┬а18 = (3├Ч1) ├Ч (3├Ч2) ├Ч (3├Ч3) ├Ч (3├Ч4) ├Ч(3├Ч5) ├Ч (3├Ч6)

= 36┬а(1├Ч2├Ч3├Ч4├Ч5├Ч6)

= 36┬а(6!)

(iii)┬а(n + 1) (n + 2) (n +3) тАж (2n)

Let us evaluate

(n + 1) (n + 2) (n +3) тАж (2n) = [(1) (2) (3) .. (n) тАж (n + 1) (n + 2) (n + 3) тАж (2n)] / (1) (2) (3).. (n)

= (2n)!/n!

(iv)┬а1┬атЛЕ┬а3┬атЛЕ┬а5┬атЛЕ┬а7┬атЛЕ┬а9 тАж (2n тАУ 1)

Let us evaluate

1┬атЛЕ┬а3┬атЛЕ┬а5┬атЛЕ┬а7┬атЛЕ┬а9 тАж (2n тАУ 1) = [(1) (3) (5) тАж (2n-1)][(2) (4) (6) тАж (2n)] / [(2) (4) (6) тАж (2n)]

= [(1) (2) (3) (4) тАж(2n-1) (2n)] / 2n┬а[(1) (2) (3) тАж (n)]

= (2n)! / 2n┬аn!

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