MCQ on Permutations and Combinations for competitive exams like Bank PO/Clerk, SBI PO, SBI Clerk, IBPS PO, IBPS Clerk, IBPS RRB, RBI, SSC, SSC CGL, SSC CHSL, SSC MTS, SSC GD, RRB NTPC, RRB Group D, AFCAT, CDS, CLAT, CAPF, and NDA.
In how many different ways can the letters of the word ‘SECOND be arranged?
(a) 720
(b) 120
(c) 5040
(d) 270
(e) None of these
Solution: (a)
The word SECOND consists of 6 distinct letters.
Therefore, Number of arrangements = \(\displaystyle 6!\)
= \(\displaystyle 6\times 5\times 4\times 3\times 2\times 1=720\)
In how many different ways can the letters of the word ‘SIMPLE’ be arranged?
(a) 520
(b) 120
(c) 5040
(d) 270
(e) None of these
Solution: (e)
The word SIMPLE consists of 6 distinct letters.
Therefore, Number of arrangements = \(\displaystyle 6!\)
= \(\displaystyle 6\times 5\times 4\times 3\times 2\times 1=720\)
In how many different ways can the letters of the word ‘BELIEVE’ be arranged?
(a) 840
(b) 1680
(c) 2520
(d) 5040
(e) None of these
Solution: (a)
The word BELIEVE consists of 7 letters in which E comes thrice.
Therefore, Required number of arrangements = \(\displaystyle \frac{{7!}}{{3!}}=\frac{{7\times 6\times 5\times 4\times 3\times 2\times 1}}{{3\times 2\times 1}}=840\)
In how many different ways can the letters of the word ‘MARKERS’ be arranged?
(a) 840
(b) 5040
(c) 2520
(d) 1680
(e) None of these
Solution: (c)
The word “MARKERS” has seven letters, and seven letters can be arranged in 7 ways.
But the letter ‘R’ appears twice.
Therefore, The number of possible ways = \(\displaystyle \frac{{7!}}{{2!}}=2520\)
Which of the following words can be written in 120 different ways?
(a) STABLE
(b) STILL
(c) WATER
(d) NOD
(e) DARE
Solution: (c)
(1) The word STABLE has six distinct letters.
Therefore, Number of arrangements = \(\displaystyle 6!\)
\(\displaystyle 6\times 5\times 4\times 3\times 2\times 1=720\)
(2) The word STILL has five letters in which letter ‘L’ comes twice.
Therefore Number of arrangements = \(\displaystyle \frac{{5!}}{2}=60\)
(3) The word WATER has five distinct letters.
Therefore Number of arrangements = \(\displaystyle 5!=5\times 4\times 3\times 2\times 1=120\)
(4) The word ‘NOD’ has 3 distinct letters.
Therefore Number of arrangements = \(\displaystyle 3!=6\)
(5) Number of arrangements = \(\displaystyle 4!=24\)
