Fill In The Blanks[1 Marks ]

Question 11 Mark

The number of permutations of n different objects$,$ taken $r$ at a line$,$ when repetitions are allowed$,$ is $....$

Answer

The number of permutations of $n$ different objects, taken $r$ at a line, when repetitions are allowed, is $=n^r$
Number of permutation of $n$ different things takan $r$ at a time when repetition is allowed $=$ filling $r$ pleces with the help of $n$ different objects when repetition ia alowed $= n \times n \times n ...r \times = n^r$

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Question 21 Mark

If $\ ^nP_r= 840, \ ^nC_r = 35,$ then $r =......$

Answer

If $\ ^nP_r= 840, ^nC_r = 35, $then $r = 4$
We have$, \ ^nP_r = 840$ and $^nC_r = 35$
Now $\ ^nP_r = \ ^nC_r . r!$
$\Rightarrow840=35 \times\text{r!}\ \Rightarrow\ ^.\text{r}!=24$
$\therefore\text{r}=4$

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Question 31 Mark

A committee of $6$ is to be chosen from $10$ men and $7$ women so as to contain atleast $3$ men and $2$ women. In how many different ways can this be done if two particular women refuse to serve on the same committee.
$[$Hint: At least $3$ men and $2$ women: The number of ways $= \ ^{10}C_3\times \ ^7C_3 + \ ^{10}C_4 \times \ ^7C_2 .$ For $2$ particular women to be always there: the number of ways $= \ ^{10}C4 + \ ^{10}C_3\times \ ^5C_1 $. The total number of committees when two particular women are never together$ \ =$ Total $–$ together$.]$

Answer

Number of men $= 10$ and number of women $= 7$
We have to from a committee of 6 persons containiing at least $3$ men and $2$ women containing at least $3$ men and $2$ women.
$\therefore$ Number of ways $=\ ^{10}\text{C}_3\times\ ^7\text{C}_3+\ ^{10}\text{C}_4\times\ ^7\text{C}_2=120\times35+210\times21=8610$
Total no. of committee of wats $-$ Number of ways when two particular women are together.
$=(\ ^{10}\text{C}_3\times\ ^7\text{C}_3+\ ^{10}\text{C}_4\times\ ^7\text{C}_2)-(\ ^{10}\text{C}_4+\ ^{10}\text{C}_3\times\ ^5\text{C}_1)$
$=(120\times35+210\times21)-(210+120\times5)$
$=4200+4410-(210+600)=7800$

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Question 41 Mark

$^{15}C_8 +^{15}C_9 – ^{15}C_6 – ^{15}C_7 = ......$

Answer

$\ ^{15}C_8 +\ ^{15}C_9 – \ ^{15}C_6 – \ ^{15}C_7 = 0$
$^{15}\text{C}_8+\ ^{15}\text{C}_9-\ ^{15}\text{C}_6-\ ^{15}\text{C}_7=\ ^{15}\text{C}_8+\ ^{15}\text{C}_9-\ ^{15}\text{C}_{15-6}-\ ^{15}\text{C}_{15-7}$ $[\because\ ^\text{n}\text{C}_\text{r}=\ ^\text{n}\text{C}_\text{n-r}]$
$=\ ^{15}\text{C}_8+\ ^{15}\text{C}_9-\ ^{15}\text{C}_9-\ ^{15}\text{C}_8=0$

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Question 51 Mark

Three balls are drawn from a bag containing $5$ red$, 4$ white and $3$ black balls. The number of ways in which this can be done if at least $2$ are red is$.....$

Answer

Three balls are drawn from a bag containing $5$ red$, 4$ white and $3$ black balls. The number of ways in which this can be done if at least $2$ are red is $= 80$
The possoble selection can be either ' two red and one other then red ' or ' three red '.
So$,$ the required number of ways $= \ ^5C_2 \times \ ^7C_1 + \ ^5C_3 = 10 \times 7 + 10 = 80$

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Question 61 Mark

The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______.
[Hint: Number of ways of arranging 6 consonants of which two are alike is $\frac{6!}{2!}$ and number of ways of arranging vowels $=\ ^7\text{P}_6\times\frac{1}{3!}\times\frac{1}{2!}.$ ]

Answer

The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is = 151200
Solution:
Letters of the word INTERMEDIATE are: vowels (I, E, E, I, A, E) and consonants (N, T, R, M, D, T)
Now we have to arrnge these letters if no tow vowels come together.
So, first arrange six consonants in $\frac{61}{2!}$ ways.
Arrangment of six consonants creates seven gaps.
Six vowels can be arranged in these gaps in $^7\text{C}_6\times\frac{6!}{2!\ 3!}$ ways.
So, total number of words $=\frac{6!}{2!}\times\ ^7\text{C}_6\times\frac{6!}{2!\ 3!}=360\times7\times60=151200$

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Question 71 Mark

In a football championship$, 153$ matches were played . Every two teams played one match with each other. The number of teams$,$ participating in the championship is $......$

Answer

In a football championship,$ 153$ matches were played . Every two teams played one match with each other. The number of teams, participating in the championship is $= 18$
Let the number of them participating in chapionship be $n.$
It is given thet every two teams played one mathch with each other.
$\therefore $ Total mathch played $= ^nC_2$
$\Rightarrow\ ^\text{n}\text{C}_2=153$
$\Rightarrow\frac{\text{n}(\text{n}-1)}{2}=153$
$\Rightarrow\text{n}^2-\text{n}-306=0$
$\Rightarrow(\text{n}-18)(\text{n}+17)=0$
$\therefore\ \text{n}=18$

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Question 81 Mark

A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.

Answer

A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is = 64
Solution:
There are 2 white, 3 black and 4 red balls.
Three balls are drawn and it is given that at least one black ball is to be included in the draw.
$\therefore$ Required number of ways $=\ ^3\text{C}_1\times\ ^6\text{C}_2+\ ^3\text{C}_2\times\ ^6\text{C}_1+\ ^3\text{C}_3$
$= 3\times15+3\times6+1=45+18+1=64$

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Question 91 Mark

The number of six$-$digit numbers$,$ all digits of which are odd is$ .....$

Answer

The number of six$-$digit numbers, all digits of which are odd is $= 5^6$
Odd digits are$: 1, 3, 5, 7, 9$
$\therefore$ The Required number of number $= 5 \times 5\times 5 \times 5 \times 5 \times 5 = 5^6$

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Question 101 Mark

The total number of ways in which six $'\ +’$ and four $'\ –\ ’$ signs can be arranged in a line such that no two signs $'\ –\ ’$ occur together is $....$

Answer

The total number of ways in which six $'\ +\ ’$ and four $'\ –\ ’ $signs can be arranged in a line such that no two signs $'\ –\ ’$ occur together is $= 35$
First arrange six $'\ +\ '$ simgs.
Number of ways of arrangment is $'\ 1\ ' ($as sings are identical$).$
Now$,$ sevan gaps are created fore which fore are to be chosen to put four $'\ - n\ '$sings.
Fore gaps can be selected in $^7C_4$ whay.
In these four gaps$, '\ - \ '$ can be arranged in only one way as sings are identical.
$\therefore$ Total number of ways $= 1 \times ^7C_4 \times 1 = 35$

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