Quantitative Aptitude
PERMUTATION AND COMBINATION MCQs
Permutations And Combinations
Total Questions : 444
| Page 40 of 45 pages
Answer: Option C. -> 720
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.
Therefore Required number of ways = (120 x 6) = 720
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.
Therefore Required number of ways = (120 x 6) = 720
Answer: Option D. -> 50400
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 = $$\frac{{7!}}{{2!}}$$ = 2520
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in $$\frac{{5!}}{{3!}}$$ = 20 ways
∴ Required number of ways = (2520 x 20) = 50400
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 = $$\frac{{7!}}{{2!}}$$ = 2520
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in $$\frac{{5!}}{{3!}}$$ = 20 ways
∴ Required number of ways = (2520 x 20) = 50400
Answer: Option C. -> 360
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
∴ Required number of ways
$$ = \frac{{6!}}{{\left( {1!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}} = 360$$
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
∴ Required number of ways
$$ = \frac{{6!}}{{\left( {1!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}} = 360$$
Answer: Option B. -> 70
Required number of ways
$$\eqalign{
& = {}^8{C_4} \cr
& = \frac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1}} \cr
& = 70 \cr} $$
Required number of ways
$$\eqalign{
& = {}^8{C_4} \cr
& = \frac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1}} \cr
& = 70 \cr} $$
Answer: Option C. -> 120
The given word contains 5 letters, all different.
∴ Required number of ways
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
The given word contains 5 letters, all different.
∴ Required number of ways
= 5!
= 5 × 4 × 3 × 2 × 1
= 120
Answer: Option C. -> 72
Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters.
These letters can be arranged in 4! = 24 ways
The letters EA may be arranged amongst themselves in 2 ways.
Number of arrangements having vowels together = (24 × 2) = 48 ways
Total arrangements of all letters
= 5!
= (5 × 4 × 3 × 2 × 1)
= 120
Number of arrangements not having vowels together
= (120 - 48)
= 72
Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters.
These letters can be arranged in 4! = 24 ways
The letters EA may be arranged amongst themselves in 2 ways.
Number of arrangements having vowels together = (24 × 2) = 48 ways
Total arrangements of all letters
= 5!
= (5 × 4 × 3 × 2 × 1)
= 120
Number of arrangements not having vowels together
= (120 - 48)
= 72
Answer: Option A. -> 2520
The given word contains 7 letters of which F is taken 2 times.
∴ Required number of ways
$$\eqalign{
& = \frac{{7!}}{{2!}} \cr
& = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr
& = 2520 \cr} $$
The given word contains 7 letters of which F is taken 2 times.
∴ Required number of ways
$$\eqalign{
& = \frac{{7!}}{{2!}} \cr
& = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr
& = 2520 \cr} $$
Answer: Option C. -> 720
The letters of the word 'BAKERY' be arranged in 6! ways
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
The letters of the word 'BAKERY' be arranged in 6! ways
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
Answer: Option C. -> 1023
GIVEN THAT, THE QUESTION PAPER CONSISTS OF FIVE PROBLEMS. FOR EACH PROBLEM, ONE OR TWO OR THREE OR NONE OF THE CHOICES CAN BE ATTEMPTED.
HENCE, THE REQUIRED NUMBER OF WAYS = 45 – 1.
= 210 – 1 = 1024 – 1 = 1023
GIVEN THAT, THE QUESTION PAPER CONSISTS OF FIVE PROBLEMS. FOR EACH PROBLEM, ONE OR TWO OR THREE OR NONE OF THE CHOICES CAN BE ATTEMPTED.
HENCE, THE REQUIRED NUMBER OF WAYS = 45 – 1.
= 210 – 1 = 1024 – 1 = 1023
Answer: Option B. -> 6! * ⁷P₆
WE CAN INITIALLY ARRANGE THE SIX BOYS IN 6! WAYS.
HAVING DONE THIS, NOW THREE ARE SEVEN PLACES AND SIX GIRLS TO BE ARRANGED. THIS CAN BE DONE IN ⁷P₆ WAYS.
HENCE REQUIRED NUMBER OF WAYS = 6! * ⁷P₆
WE CAN INITIALLY ARRANGE THE SIX BOYS IN 6! WAYS.
HAVING DONE THIS, NOW THREE ARE SEVEN PLACES AND SIX GIRLS TO BE ARRANGED. THIS CAN BE DONE IN ⁷P₆ WAYS.
HENCE REQUIRED NUMBER OF WAYS = 6! * ⁷P₆
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