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Quantitative Ability Tech Test Permutation and Combinations Test 1

2889.In how many different ways can the letters of the word JUDGE be arranged such that the vowels always come together?
None of these
48
32
64
Explanation:

The word JUDGE has 5 letters. It has 2 vowels [UE] and these 2 vowels should always come together. Hence these 2 vowels can be grouped and considered as a single letter. That is, JDG[UE].

Hence we can assume total letters as 4 and all these letters are different. Number of ways to arrange these letters

= 4!=4×3×2×1=24

In the 2 vowels [UE], all the vowels are different. Number of ways to arrange these vowels among themselves

=2!=2×1=2

Total number of ways =24×2=48

2893.A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 8 from part P and 4 from part Q, in how many ways can he do that?
None of these
6020
1200
9450
Explanation:

Number of ways to choose 8 questions from part P = 10C8

Number of ways to choose 4 questions from part Q = 10C4

Total number of ways

= 10C8 × 10C4

= 10C2 × 10C4 [nCr = nC(n-r)]

=$\left(\dfrac{10 \times 9}{2 \times 1}\right) \left(\dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}\right)$=45×210=9450

2895.In how many different ways can the letters of the word CORPORATION be arranged so that the vowels always come together?
810
1440
2880
50400
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 =$ \dfrac{7!}{2!} $= 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in$ \dfrac{5!}{3!} $= 20 ways.

$\therefore$ Required number of ways = $\left(2520 \times 20\right)$ = 50400.

2896.In how many different ways can the letters of the word MATHEMATICS be arranged so that the vowels always come together?
10080
4989600
120960
None of these
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.

$\therefore$ Number of ways of arranging these letters =$ \dfrac{8!}{(2!)(2!)} $= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =$ \dfrac{4!}{2!} $= 12.

$\therefore$ Required number of words =$\left (10080 \times 12\right)$ = 120960.

44366.In how many different ways can the letters of the word 'RUMOUR' be arranged?
128
130
180
200
Explanation:

The word 'RUMOUR' has 6 letters.

In these 6 letters, 'R' occurs 2 times, 'U' occurs 2 times and rest of the letters are different.

Hence, number of ways to arrange these letters

=$\dfrac{6!}{\left(2!\right)\left(2!\right)}$

=$\dfrac{6\times5\times4\times3\times2\times1}{\left(2\times1\right)\left(2\times1\right)}$

=180

44367.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
1200
1800
1900
Explanation:

5 subjects can be arranged in 6 periods in $^6P_{5}$ ways.

Any of the 5 subjects can be organized in the remaining period ( $^5C_{1}$ ways).

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

Total number of arrangements

$\dfrac{^6P_{5}\times^5C_{1}}{\left(2!\right)}$

=1800

44368.In a chess competition involving some men and women, every player needs to play exactly one game with every other player. It was found that in 45 games, both the players were women and in 190 games, both players were men. What is the number of games in which one person was a man and other person was a woman?
40
80
200
120
Explanation:

Let total number of women =w

total number of men =m

Number of games in which both players were women =45

=>$wc_2 $= 45

=>$\dfrac{w(w-1)}{2}$=45

=>w(w-1)=90

=>w=10

Number of games in which both players were men =190

=>mc_2 = 190

=>$\dfrac{m(m-1)}{2}$=190

=>m(m-1)=380

=>m=20

We have got that

Total number of women = 10

Total number of men = 20

Required number of games in which one person was a man and other person was a woman

$^{20}C_1 × ^{10}C_1$ =20×10=200

44369.In how many ways can 11 persons be arranged in a row such that 3 particular persons should always be together?
9!×3!
9!
11!
11!×3!
Explanation:

Given that three particular persons should always be together. Hence, just group these three persons together and consider as a single person.

Therefore we can take total number of persons as 9. These 9 persons can be arranged in 9!ways.

We had grouped three persons together. These three persons can be arranged among themselves in 3!ways.

Hence, required number of ways

=9!×3!

44370.A company has 11 software engineers and 7 civil engineers. In how many ways can they be seated in a row so that all the civil engineers do not sit together?
$^{18}P_{4} - 2!$
18! - (12! × 7!)
$^{18}P_{4} × 11$
18!- (11! × 7!)
Explanation:

Total number of engineers =11+7=18.

These 18 engineers can be arranged in a row in 18!ways. ...(A)

Now we will find out the total number of ways in which these 18 engineers can be arranged so that all the 7 civil engineers will always sit together.

For this, group all the 7 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 12. These 12 engineers can be arranged in 12!ways.

We had grouped 7 civil engineers. These 7 civil engineers can be arranged among themselves in 7!ways.

Hence, total number of ways in which the 18 engineers can be arranged so that the 7 civil engineers will always sit together = 12! x 7! ...(B)

From (A) and (B),

Total number of ways in which 11 software engineers and 7 civil engineers can be seated in a row so that all the civil engineers do not sit together=18!-(12! x7!)

44371.How many signals can be made using 6 different coloured flags when any number of them can be hoisted at a time?
1956
1720
2020
1822
Explanation:

Given that any number of flags can be hoisted at a time. Hence we need to find out number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.

Number of signals that can be made using 1 flag

= $^6P_{1}$ =6

Number of signals that can be made using 2 flags

=$ ^6P_{2} $

=6×5=30

Number of signals that can be made using 3 flags

=$ ^6P_{3}$

=6×5×4=120

Number of signals that can be made using 4 flags

=$ ^6P_{4} $

=6×5×4×3=360

Number of signals that can be made using 5 flags

= $^6P_{5}$

=6×5×4×3×2=720

Number of signals that can be made using 6 flags

= $^6P_{6 }$

=6×5×4×3×2×1=720

Therefore, required number of signals

=6+30+120+360+720+720=1956

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