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How many different four-letter words can be formed (the wor
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14 May 2019, 00:05
14 May 2019, 00:05
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46% (02:04) correct
53% (01:40) wrong based on 43 sessions
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How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS such that each word starts with G and ends with R?
(A) \(8P_2\)
(B) \(\frac{8P_2}{2!*2!}\)
(C) \(8P_4\)
(D) \(\frac{8P_4}{2!*2!}\)
(E) \(\frac{10P_2}{2!*2!}\)
(A) \(8P_2\)
(B) \(\frac{8P_2}{2!*2!}\)
(C) \(8P_4\)
(D) \(\frac{8P_4}{2!*2!}\)
(E) \(\frac{10P_2}{2!*2!}\)
Re: How many different four-letter words can be formed (the wor
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10 Feb 2020, 19:30
10 Feb 2020, 19:30
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Carcass wrote:
How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS such that each word starts with G and ends with R?
(A) \(8P_2\)
(B) \(\frac{8P_2}{2!*2!}\)
(C) \(8P_4\)
(D) \(\frac{8P_4}{2!*2!}\)
(E) \(\frac{10P_2}{2!*2!}\)
(A) \(8P_2\)
(B) \(\frac{8P_2}{2!*2!}\)
(C) \(8P_4\)
(D) \(\frac{8P_4}{2!*2!}\)
(E) \(\frac{10P_2}{2!*2!}\)
Place one G in the first slot and one R in the last slot:
G __ __ R
The remaining letters, {G, R, E, A, I, O, U, S}, can be arranged in the remaining 2 slots in \(8P_2\) [no indistinguishable(same) objects nor repetition]. The answer is (A).
Note: Since the two G’s in the base word are indistinguishable, the word G 1 G 2 AR is the same as G 2 G 1 AR. Hence, the internal arrangement of the G’s or, for the same reason, the R’s is not important.
Re: How many different four-letter words can be formed (the wor
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18 Aug 2020, 11:03
18 Aug 2020, 11:03
Can you please explain how they are not important?
To me, G1 G2 R1 R2 are the same, so we would get repetition with GGRR, GRGR, GG_R, and G_RR
To me, G1 G2 R1 R2 are the same, so we would get repetition with GGRR, GRGR, GG_R, and G_RR
