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Statistics 2 Mark Each

Permutations , combinations and binomial expansion · Gujarat Board (GSEB) STD 11 Commerce (GSEB - English Medium). 60 questions.

Questions · Page 2 of 2

2 Mark Each

Question 512 Marks
Using all the digits $2, 3, 5, 8, 9$, how many numbers greater than $50,000$ can be formed ?
Answer
Given digits are $2,3,5,8,9$
For the numbers greater than $50.000$ the digit at the first place may be $5$ or greater than It.
$\therefore$ Out of the digit $5,8,9$, the first digit can be placed in ${ }^3 P_1$ ways.
Now, from the remalning $4$ dlgits, all four can be placed in $4 P 4$ ways.
$\therefore$ Total permutations for the numbers greater than $50,000={ }^3 P_1 \times{ }^4 P_4$
$=3 \times 4!$
$=3 \times 24$
$=72$
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Question 522 Marks
There are $7$ cages for $7$ lions in a zoo. $3$ cages out of $7$ cages are so small that $3$ out of $7$ lions cannot fit in it. In how many ways can $7$ lions be caged in $7$ cages ?
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Answer
Except 3 small cages In remaining $4$ cages $3$ lions can be caged $\ln { }^4 P_3$ ways.
Now, In the remainlng ( $1$ big $+3$ small) $4$ cages the remaining $4$ lions can be caged $\ln { }^4 \mathrm{P}_4$ ways.
$\therefore$ Total permutations of caging $7$ lions in
$7 \text { cages }={ }^4 P_3 \cdot{ }^4 P_4$
$=(4 \times 3 \times 2) \times 4!$
$=24 \times 24$
$=576$
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Question 532 Marks
In how many ways can $5$ boys and $3$ girls be arranged In a row such that all the boys are together ?
Answer
$5$ boys and $3$ gIrls are to be arranged In a row such that all the boys are together.
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$5$ boys are to be arranged together. therefore considering them as one person total $4$ persons can be arranged In ${ }^4 P_4$ ways.
Now. In each of these arrangments $5$ boys can be arranged among themselves in ${ }^5 P_5$ ways.
$\therefore \text { Total permutation }={ }^4 \mathrm{P}_4 \times{ }^5 \mathrm{P}_5$
$=4!\times 5!$
$=24 \times 120$
$=2880$
Hence, $5$ boys and $3$ girls can be arranged In a row such that $5$ boys are together in $2880$ ways.
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Question 542 Marks
How many six digit numbers can be formed using all the digits $1,2,3,0,7,9$ ?
Answer
Using all the digits $1,2,3,0,7,9$, six digit numbers are to be formed.
$\therefore$ Excluding digit $0$ , one of the five digits can be placed at the first place in ${ }^5 P_1$ ways Now, remaining $5$ digits (including $0$ ) can be arranged in remaining $5$ places in ${ }^5 P_5$ ways.
$\therefore$ Total permutations for six digit numbers
$={ }^5 \mathrm{P}_1 \times{ }^5 \mathrm{P}_5$
$=5 \times 5!$
$=5 \times 120$
$=600$
Hence, $600$ numbers of $6$ digit can be formed.
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Question 552 Marks
If ${ }^{9} P_{r}=3024$, find the value of $r$.
Answer
${ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}=\frac{n !}{(n-r) !}$
$ { }^9 \mathrm{P}_{\mathrm{r}}=\frac{9 !}{(9-r) !}$
$ \therefore 3024=\frac{362880}{(9-r) !}$
$ \therefore(9-r) !=\frac{362880}{3024}$
$ \therefore(9-r) !=120=5 !$
$ \therefore 9-\mathrm{r}=5$
$ \therefore \mathrm{r}=9-5=4$
Hence, $r=4$
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Question 562 Marks
There are $5$ seats in a car including the driver’s seat. If $3$ out of $10$ members in a family know driving then in how many ways, $5$ persons out of $10$ members can be arranged in the car ?
Answer
$3$ out of $10$ members in a family know driving.
$\therefore$ On the driver's seat a member can be place in ${ }^3 \mathrm{P}_1$ ways.
Now, there are $5$ seats in a car including the driver's seat.
The remaining $4$ members out of remaining $9$ members of the family can be placed in ${ }^9 P_4$ ways.
$\therefore \text { Total permutations of such arrangement }$
$={ }^3 \mathrm{P}_1 \times{ }^9 \mathrm{P}_4$
$=3 \times(9 \times 8 \times 7 \times 6)$
$=3 \times 3024$
$=9072$
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Question 572 Marks
What is the ratio of number of arrangements of all letters of the word $ASHOK$ and $GEETA ?$
Answer
Total permutations of arranging $5$ letters of word $ASHOK ={ }^5 P_5=5 !=120$
Total permutations of arranging $5$ letters of which $E$ is repeated $2 times$ of word $GEETA =\frac{5 !}{2 !}=\frac{120}{2}=60$
$\therefore$ The ratio of number of arrangements of all letters of the word $ASHOK$ and $GEETA =\frac{120}{2}=2: 1$
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Question 582 Marks
A person has $5$ chocolates of different sizes. These chocolates are to be distributed among $5$ children of different ages. If the biggest chocolate is to be given to the youngest child then in how many ways, $5$ chocolates can be distributed among $5$ children $?$
Answer
A person has $5$ chocolates of different sizes and they are to be distributes among $5$ children of different ages. If the biggest chocolate is to be given to the youngest child, then remaining $4$ chocolates can be distributed among the remaining $4$ children in ${ }^4 P_4$ ways.
$\therefore \text { Total permutations }={ }^1 \mathrm{P}_1 \times{ }^4 \mathrm{P}_4$
$=1 \times 4!$
$=1 \times 24$
$=24$
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Question 602 Marks
Obtain the expansion of following binomial expressions: $(3 a+4 b)^{3}$
Answer
$27 a^{3}+108 a^{2} b+144 a b^{2}+64 b^{3}$
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2 Mark Each - Page 2 - Statistics STD 11 Commerce Questions - Vidyadip