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Quantitative Ability Tech Test Permutation and Combinations Test 2

2874.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?
564
645
735
756
Explanation:

We may have [3 men and 2 women] or [4 men and 1 woman] or [5 men only].

$\therefore$ Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times\dfrac{6 \times 5}{2 \times 1} \right) $+ (7C3 x 6C1) + (7C2)
= 525 +$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 6\right) $+$ \left(\dfrac{7 \times 6}{2 \times 1} \right) $
= (525 + 210 + 21)
= 756.
2875.In how many different ways can the letters of the word OPTICAL be arranged so that the vowels always come together?
120
720
4320
2160
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.

$\therefore$ Required number of ways = $\left(120 \times 6\right)$ = 720.

2876.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?
159
194
205
209
Explanation:

We may have [1 boy and 3 girls] or [2 boys and 2 girls] or [3 boys and 1 girl] or [4 boys].

$\therefore$ Required number
of ways		 = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
=$\left (6 \times 4\right)$ +$ \left(\dfrac{6 \times 5}{2 \times 1} \times\dfrac{4 \times 3}{2 \times 1} \right) $+$ \left(\dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4\right) $+$ \left(\dfrac{6 \times 5}{2 \times 1} \right) $
= (24 + 90 + 80 + 15)
= 209.
2877.There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?
3200
None of these
1800
3600
Explanation:

5 subjects can be arranged in 6 periods in 6P5 ways.

Any of the 5 subjects can be organized in the remaining period 5C1 ways.

Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.

Total number of arrangements

$=\dfrac{~^{6}P_5 × ~^5C_1}{2!}=1800$

2878.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?
63
90
126
45
Explanation:
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =$ \left(\dfrac{7 \times 6}{2 \times 1} \times 3\right) $= 63.
2879.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?
32
48
36
60
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 = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = $\left(6 \times 6\right)$ = 36.

2882.Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
210
1050
25200
21400
Explanation:

Number of ways of selecting [3 consonants out of 7] and [2 vowels out of 4]

      = (7C3 x 4C2)
=$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times\dfrac{4 \times 3}{2 \times 1} \right) $
= 210.

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.

$\therefore$ Required number of ways = $\left(210 \times 120\right)$ = 25200.

2885.In how many different ways can the letters of the word LEADING be arranged in such a way that the vowels always come together?
360
480
720
5040
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.

$\therefore$ Required number of ways = $\left(120 \times 6\right)$ = 720.

2887.In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate?
2880
1400
1200
3212
Explanation:

Around a circle, 5 boys can be arranged in 4! ways.

Given that the boys and the girls alternate. Hence there are 5 places for the girls. Therefore the girls can be arranged in 5! ways.

Total number of ways

=4!×5!=24×120=2880

2888.How many words can be formed by using all letters of the word BIHAR?
720
24
120
60
Explanation:

The word BIHAR has 5 letters and all these 5 letters are different.

Total number of words that can be formed by using all these 5 letters

= 5P5=5

=5×4×3×2×1=120

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