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Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.
36
48
144
96
Answer with explanation
Answer: Option CExplanation
Explanation: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.
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Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition.
666600
665500
666700
678860
Answer with explanation
Answer: Option AExplanation
Explanation: The given digits are 1, 3, 5, 7, 9
Sum of r digit number= n-1Pr-1
(Sum of all n digits)×(1111… r times)
N is the number of non zero digits.
Here n=5, r=4
The sum of 4 digit numbers
4P3 (1+3+5+7+9)(1111)=666600.
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