4 alphabets E, K, S and V,one in each,were pu | vedclass

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(B) The problem asks for the number of ordered pairs that can be formed from $4$ distinct alphabets $(E, K, S, V)$.Since the order of the alphabets ma

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$12$

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(B) The problem asks for the number of ordered pairs that can be formed from $4$ distinct alphabets $(E, K, S, V)$.Since the order of the alphabets matters (as they are used as initials),we use the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.Here,$n = 4$ (total alphabets) and $r = 2$ (number of alphabets to be chosen for the pair).Number of ordered pairs $= P(4, 2) = \frac{4!}{(4-2)!} = \frac{4 	imes 3 	imes 2 	imes 1}{2 	imes 1} = 4 	imes 3 = 12$.Therefore,the total number of ordered pairs is $12$.

$4$ alphabets $E, K, S$ and $V$,one in each,were purchased from a plastic warehouse. How many ordered pairs of alphabets,to be used as initials,can be formed from them?

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