11In how many ways can the letters of the word 'BEAUTY' be arranged so that the words always start with $A$ and does not ends with $B$? A96 B78 C40 D36
Answer: A
Explanation:
$BEAUTY$ contains $6$ letters.
Start with $A$ and Does not end with $B$ $=$ (All words start with $A$) $-$ (All words start with $A$ and Ends with $B$)
If we fix the first letter $A$ in the first place,
$\underline A \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} $
Then number of ways of filling Remaining $5$ letters in $5$ places $= 5! = 120$
If we fix the first letter $A$ in the first place and $B$ in the last place,
$\underline A \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline B $
Then number of ways of filling Remaining $4$ letters in $4$ places $= 4! = 24$
Start with $A$ and Does not end with $B$ $= 120 - 24 = 96$
12In how many ways can the letters of the word 'BEAUTY' be arranged so that the vowels are always together? A112 B128 C144 D172
Answer: C
Explanation:
Vowels in the word are $AEU$, consonents are $BTY$
Put all vowels in a box and name it $X$.
$\boxed{AEU}_{\rightarrow{X}}, B, T, Y $
Number of ways of arranging above $4$ letters $=4! = 24$
Now, the number of ways the vowels in that box arrange themselves $=3! = 6$
Total ways = $24 \times 6 = 144$
13
How many different words with or without meaning can be formed using all the letters of the word 'GORGEOUS'?
A$8!$ B$15400$ C$10080$ D$\dfrac{{8!}}{{2!}}$
Answer: C
Explanation: Formula:Number of ways of arranging $n$ items of which $p$ are alike, $q$ alike and so on $ = \dfrac{{n!}}{{p! \times q! \times ...}}$
$GORGEOUS$ $= (GG), (OO), R, E, U, S$
The required number of words formed from all the letters of the given word $ = \dfrac{{8!}}{{2! \times 2!}}$ $= 10080$
14
From a group of 12 persons, in how many ways can a selection of 5 persons be made?
A1280 B625 C680 D792
Answer: D
Explanation:
Number of ways of selecting $5$ persons out of $12$ persons $={}^{12}{C_5}$ $ = \dfrac{{12 \times 11 \times 10 \times 9 \times 8}}{{5!}}$$ = \dfrac{{12 \times 11 \times 10 \times 9 \times 8}}{{5 \times 4 \times 3 \times 2 \times 1}}$ $=792$
15
From a group of 12 persons, in how many ways can a selection of 5 persons be made such that, A particular person is always included?
A150 B225 C250 D330
Answer: D
Explanation:
If a particular prsons is already included, then we have to select $4$ persons from the remaining $11$.
$\therefore {}^{11}{C_4} = \dfrac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}} = 330$
16
From a group of 12 persons, in how many ways can a selection of 5 persons be made such that, A particular person is always excluded?
A410 B460 C462 D450
Answer: C
Explanation:
If a particular prsons is always excluded, then we have to select $5$ persons from the remaining $11$.
$\therefore {}^{11}{C_5} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7}}{{5 \times 4 \times 3 \times 2 \times 1}} = 462$
17
From 8 Americans and 5 Chinese , a committee of 6 is to be formed. In how many ways can it be done, If the committee consists exactly 3 chinese?
A560 B640 C582 D552
Answer: A
Explanation:
We have to choose 3 out of 5 chinese and 3 out of 8 Americans.
$\therefore {}^5{C_3} \times {}^8{C_3} $$ = 10 \times 56 = 560$
18
From 8 Americans and 5 Chinese , a committee of 6 is to be formed. In how many ways can it be done, If the committee consist at least 3 Americans?
19City P is connected to city Q with 5 diffrent highways and city Q is connected to City R with 8 different highways.In how many ways one can choose a different route to travel from P to R via city Q.
A30 B50 C40 D70
Answer: C
Explanation:
$P\xrightarrow{\text{8 routes}}Q\xrightarrow{\text{5 routes}}R$
Required number of different routes $ = {}^8{C_1} \times {}^5{C_1} = 40$
208 girls are to be made to stand in a row for a photograph. Among them three particular girls want to be together. In how many ways they can be arranged?
Explanation:
$G,G,G,G,G,\boxed{G_1,G_2,G_3}_X$
Let us group those three girls and name that group $X$
Now total number of objects $= 6$.
Number of ways of arranging Six objects $= 6!$
Now, the three girls can arrange themselves in $= 3!$ ways.
Total number of ways $= 6! \times 3!$
