Commercial Bank of Ethiopia- CBE exam
1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3Â men are there on the committee. In how many ways can it be done?
A. 564Â Â Â Â Â B. 645Â Â Â Â Â C. 735 Â Â Â Â Â D. 756Â Â Â Â Â Â Â Â Â E. None of these
2. In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
A. 360 B. 480Â Â Â C. 720 Â Â Â Â Â Â D. 5040Â Â Â Â E. None of these
2. Answer: Option C
Explanation:
The word ‘LEADING’ has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
3. In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
A. 810 Â Â B. 1440 Â Â C. 2880 Â Â D. 50400 Â Â E. 5760
3. Answer: Option D
Explanation:
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters = 7!/2!
= 2520. Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways. Required number of ways = (2520 x 20) = 50400.
4. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 210 B. 1050Â Â Â C. 25200Â Â Â Â Â Â D. 21400Â Â Â Â Â E. None of these
Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120. Required number of ways = (210 x 120) = 25200
5. In how many ways can the letters of the word ‘LEADER’ be arranged?
A. 72Â Â Â Â B. 144Â Â Â Â Â C. 360Â Â Â Â Â D. 720Â Â Â Â E. None of these
5. Answer: Option C
Explanation:
The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. Required number of ways = 6!/(1!)(2!)(1!)(1!)(1! = 360.
6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
A. 159 Â Â Â B. 194Â Â C. 205Â Â Â Â D. 209Â Â E. None of these
6. Answer: Option D
Explanation: We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4Â boys). Required number of ways = (6C1 x 4 C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)Â Â = (6 x 4) +(6 x 5/2 x 1×4 x 3/2 x 1)+6 x 5 x 4/3 x 2 x 1)x4)+ 6 x 5/2 x 1
= (24 + 90 + 80 + 15)
= 209
7. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
A. 5 Â Â Â Â Â Â B. 10Â Â Â C. 15Â Â Â Â Â Â Â Â Â Â D. 20
7. Answer: Option D
Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.
8. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
A. 266 Â Â Â Â Â B. 5040Â Â Â Â Â Â C. 11760Â Â Â Â Â Â Â Â Â D. 86400Â Â Â Â E. None of these
8. Answer: Option C
9. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3
balls be drawn from the box, if at least one black ball is to be included in the draw?
A. 32 Â Â Â Â B. 48Â Â C. 64 D. 96Â Â Â Â Â Â Â Â Â Â Â E. None of these
9. Answer: Option C
10. In how many different ways can the letters of the word ‘DETAIL’ be arranged in
such a way that the vowels occupy only the odd positions?
A. 32 B. 48Â Â Â Â Â Â Â C. 36Â Â Â Â Â Â Â Â Â Â Â D. 60Â Â Â Â Â Â Â Â Â E. 120
10. Answer: Option C
Explanation: There are 6 letters in the given word, out of which there are 3 vowels and 3Â consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5. Number of ways of arranging the vowels = 3 P3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3 P3 = 3! = 6. Total number of ways = (6 x 6) = 36.
11.In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A. 63 Â Â B. 90Â Â Â Â C. 126Â Â Â D. 45 Â Â E. 135
11. Answer: Option A
12. How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
A. 40Â Â Â Â B. 400Â Â Â Â C. 5040Â Â Â Â Â Â D. 2520
12. Answer: Option C
13. In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?
A. 10080Â Â Â Â B. 4989600 Â Â C. 120960Â Â Â Â D. None of these
13. Answer: Option C
Explanation:
In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Â
14. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
A. 120 Â Â Â B. 720 Â Â C. 4320Â Â D. 2160Â Â E. None of these
14.Answer: Option B
Explanation: The word ‘OPTICAL’ contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x6) = 720.
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