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
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
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.
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.
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
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
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
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!
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!)
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