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Number of arrangements of the letters of the word ENGINEER so that all vowels are together but all Es are not together is

Number of ways of arrangement the letters of the word ELEVEN so that no two vowels are together and no two consonants are together is k then 10 k equals

Number of ways of keeping 5 letters in 5 addressed envelopes such that no letter goes to its own envelope is p then p is divisible by

Number of 6 digits numbers greater than 123000 using 1,2,2,3,3,3 is

answer is A, C, D.

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Detailed Solution

A). In the word ENGINEER the vowels are E,E,E,I, consonants are N,N,G,R Required = (No. of arrangements in which E,E,E,I are together) – (No. of arrangemnets in which EEE are together when E,E,E,I are together)

$= (\frac{∠ 5}{∠ 2} \times \frac{∠ 4}{∠ 3}) - (\frac{∠ 5}{∠ 2} \times 2) = 60 \times (4 - 2) = 120 = 5!$

B). No.of arrangements of E,E,E and Consonants L,V,N alternately starting with E is $∠ 3$ and starting with a
consonant is $∠ 3$ K=Total no.of such arrangments = $∠ 3 + ∠ 3 = 12$

$\Rightarrow 10 K = 120 = 5!$

C). P = No.of Derangements of 5 letters in 5 addresed envelops $\Rightarrow P = D_{5} = 44$

44 is divisible by 22 where $\frac{1}{3} (^{12} C_{2}) = 22$

D). Using 1,2,2,3,3,3 numbers of 6 digits numbers that are starting with 123, 132,133 or 2 are greater than 123000. No.of such numbers = $\frac{∠ 3}{∠ 2} + \frac{∠ 3}{∠ 2} + \frac{∠ 3}{∠ 2} + \frac{∠ 5}{∠ 3} + \frac{∠ 5}{∠ 2 \cdot ∠ 2} = 59$