### 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 3 ****men are there on the committee. In how many ways can it be done?**

A. 564 B. 645 C. 735 D. 756 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. 480 C. 720 D. 5040 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 B. 1440 C. 2880 D. 50400 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. 1050 C. 25200 D. 21400 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. 72 B. 144 C. 360 D. 720 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 B. 194 C. 205 D. 209 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 (4 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) = (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 B. 10 C. 15 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 B. 5040 C. 11760 D. 86400 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 B. 48 C. 64 D. 96 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. 48 C. 36 D. 60 E. 120

**10. Answer: Option C****Explanation**: There are 6 letters in the given word, out of which there are 3 vowels and 3 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 B. 90 C. 126 D. 45 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. 40 B. 400 C. 5040 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. 10080 B. 4989600 C. 120960 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.

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**14. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so ****that the vowels always come together?**

A. 120 B. 720 C. 4320 D. 2160 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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