Permutation & combination Quant Quiz for SBI PO PRE 2019 (Day 23) | 27th February 2019

Permutation & combination Quant Quiz for SBI PO PRE 2019 (Day 23)

Quant Quiz to improve your Quantitative Aptitude for SBI PO & SBI clerk exam quant, IBPS PO quant , IBPS Clerk quant , IBPS RRB quant, and other competitive exams.

Q1.   In how many different ways can the letters of the word “CHARGES” be arranged in such a way that the vowels always come together

A) 1440

B) 720

C) 360

D) 240

E)  None of these 

Q2.  In how many different ways can the letters of the word “COMPLAINT” be arranged in such a way that the vowels occupy only the odd positions? 

A) 14400

B) 43200

C) 1440

D) 5420

E) None of these

Q3.  In how many different ways can the letters of the word “CANDIDATE” be arranged in such a way that the vowels always come together? 

A) 4320

B) 1440

C) 720

D) 840

E) None of these

Q4.  In how many different ways can the letters of the word “RADIUS” be arranged in such a way that the vowels occupy only the odd positions? 

A) 72

B) 144

C) 532

D) 36

E) None of these

Q5.  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

Q6.   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

Q7.   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

Q8.  In how many different ways can the letters of the word ‘SERVING’ be arranged? 

A) 5040

B) 720

C) 120

D)  40320 

E)  None of these

Q9.  In how many different ways can any 4 letters of the word ‘WORKING’ be arranged? 

A) 5040

B) 840

C) 24

D) 120

E)  None of these 

Q10.  In how many different ways can the letters of the word ‘ARRANGEMENT’ be arranged? 

A)  2494800 

B)  4989600 

C) 831600

D) Can’t be determined

E) None of these

Solutions 

Q1. Ans(A)

6! ×2!=1440

Q2. Ans(B)

Q3. Ans(A)

Q4. Ans(D)

Q5. Ans(C)

5! × 3! = 720

Q6. Ans(D)

Q7. Ans(C)

Q8. Ans(A)

Q9. Ans(B)

Since all letters are unique, choosing any four letters will lead us to 7×6×5×4 = 840 permutations.

Q10. Ans(A)

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