NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
119233
There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is
1 369600
2 400400
3 420600
4 440720
Explanation:
A Given that, there are 3 copies each of 4 different books. Total number of copies \(=12\) The desired number is \(\frac{12 !}{(3 !)^4}=\frac{479001600}{1296}=369600\)
COMEDK-2016
Permutation and Combination
119234
A library has a copies of one book, b copies of of two books, c copies of three books and single copy of \(d\) books. The total number of ways in which these books can be arranged in a shelf, is
1 \(\frac{(a+b+c+d) !}{a ! b ! c !}\)
2 \(\frac{(a+2 b+3 c+d) !}{a !(b !)^2(c !)^3}\)
3 \(\frac{(a+2 b+3 c+d) !}{a ! b ! c !}\)
4 None of these
Explanation:
B Here total number of books \(=a+2 b+3 c+d\) Number of ways they can arranged \(=(a+2 b+3 c+d)\) ! \(\therefore\) Total number of arrangements \(=\frac{(\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}+\mathrm{d}) !}{\mathrm{a} !(\mathrm{b} !)^2(\mathrm{c} !)^3}\)
COMEDK-2018
Permutation and Combination
119236
Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is
1 20
2 9
3 120
4 40
Explanation:
A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( \(\because \mathrm{H}=\) heads, \(\mathrm{T}=\) tails \()\) Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails. Thus, the required permutation is \(=\frac{6 !}{3 ! 3 !}={ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 !}=20\)
VITEEE-2019
Permutation and Combination
119240
There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is \(\qquad\)
1 120
2 \(89 .(8 !)\)
3 56
4 \(42 \times 5^4\)
Explanation:
C : First arrange 7 identical white ball in row is \(\frac{7 !}{7 !}=1\) way. Now for black balls we have 8 places from 8 places we can choose any of 3 places \(={ }^8 \mathrm{C}_3\) and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be \(\frac{3 !}{3 !}\) hence, solution should be \(\frac{7 !}{7 !} \times{ }^8 \mathrm{C}_3 \times \frac{3 !}{3 !}=56\)
119233
There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is
1 369600
2 400400
3 420600
4 440720
Explanation:
A Given that, there are 3 copies each of 4 different books. Total number of copies \(=12\) The desired number is \(\frac{12 !}{(3 !)^4}=\frac{479001600}{1296}=369600\)
COMEDK-2016
Permutation and Combination
119234
A library has a copies of one book, b copies of of two books, c copies of three books and single copy of \(d\) books. The total number of ways in which these books can be arranged in a shelf, is
1 \(\frac{(a+b+c+d) !}{a ! b ! c !}\)
2 \(\frac{(a+2 b+3 c+d) !}{a !(b !)^2(c !)^3}\)
3 \(\frac{(a+2 b+3 c+d) !}{a ! b ! c !}\)
4 None of these
Explanation:
B Here total number of books \(=a+2 b+3 c+d\) Number of ways they can arranged \(=(a+2 b+3 c+d)\) ! \(\therefore\) Total number of arrangements \(=\frac{(\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}+\mathrm{d}) !}{\mathrm{a} !(\mathrm{b} !)^2(\mathrm{c} !)^3}\)
COMEDK-2018
Permutation and Combination
119236
Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is
1 20
2 9
3 120
4 40
Explanation:
A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( \(\because \mathrm{H}=\) heads, \(\mathrm{T}=\) tails \()\) Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails. Thus, the required permutation is \(=\frac{6 !}{3 ! 3 !}={ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 !}=20\)
VITEEE-2019
Permutation and Combination
119240
There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is \(\qquad\)
1 120
2 \(89 .(8 !)\)
3 56
4 \(42 \times 5^4\)
Explanation:
C : First arrange 7 identical white ball in row is \(\frac{7 !}{7 !}=1\) way. Now for black balls we have 8 places from 8 places we can choose any of 3 places \(={ }^8 \mathrm{C}_3\) and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be \(\frac{3 !}{3 !}\) hence, solution should be \(\frac{7 !}{7 !} \times{ }^8 \mathrm{C}_3 \times \frac{3 !}{3 !}=56\)
119233
There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is
1 369600
2 400400
3 420600
4 440720
Explanation:
A Given that, there are 3 copies each of 4 different books. Total number of copies \(=12\) The desired number is \(\frac{12 !}{(3 !)^4}=\frac{479001600}{1296}=369600\)
COMEDK-2016
Permutation and Combination
119234
A library has a copies of one book, b copies of of two books, c copies of three books and single copy of \(d\) books. The total number of ways in which these books can be arranged in a shelf, is
1 \(\frac{(a+b+c+d) !}{a ! b ! c !}\)
2 \(\frac{(a+2 b+3 c+d) !}{a !(b !)^2(c !)^3}\)
3 \(\frac{(a+2 b+3 c+d) !}{a ! b ! c !}\)
4 None of these
Explanation:
B Here total number of books \(=a+2 b+3 c+d\) Number of ways they can arranged \(=(a+2 b+3 c+d)\) ! \(\therefore\) Total number of arrangements \(=\frac{(\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}+\mathrm{d}) !}{\mathrm{a} !(\mathrm{b} !)^2(\mathrm{c} !)^3}\)
COMEDK-2018
Permutation and Combination
119236
Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is
1 20
2 9
3 120
4 40
Explanation:
A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( \(\because \mathrm{H}=\) heads, \(\mathrm{T}=\) tails \()\) Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails. Thus, the required permutation is \(=\frac{6 !}{3 ! 3 !}={ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 !}=20\)
VITEEE-2019
Permutation and Combination
119240
There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is \(\qquad\)
1 120
2 \(89 .(8 !)\)
3 56
4 \(42 \times 5^4\)
Explanation:
C : First arrange 7 identical white ball in row is \(\frac{7 !}{7 !}=1\) way. Now for black balls we have 8 places from 8 places we can choose any of 3 places \(={ }^8 \mathrm{C}_3\) and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be \(\frac{3 !}{3 !}\) hence, solution should be \(\frac{7 !}{7 !} \times{ }^8 \mathrm{C}_3 \times \frac{3 !}{3 !}=56\)
119233
There are 3 copies each of 4 different books. The number of ways they can be arranged in a shelf is
1 369600
2 400400
3 420600
4 440720
Explanation:
A Given that, there are 3 copies each of 4 different books. Total number of copies \(=12\) The desired number is \(\frac{12 !}{(3 !)^4}=\frac{479001600}{1296}=369600\)
COMEDK-2016
Permutation and Combination
119234
A library has a copies of one book, b copies of of two books, c copies of three books and single copy of \(d\) books. The total number of ways in which these books can be arranged in a shelf, is
1 \(\frac{(a+b+c+d) !}{a ! b ! c !}\)
2 \(\frac{(a+2 b+3 c+d) !}{a !(b !)^2(c !)^3}\)
3 \(\frac{(a+2 b+3 c+d) !}{a ! b ! c !}\)
4 None of these
Explanation:
B Here total number of books \(=a+2 b+3 c+d\) Number of ways they can arranged \(=(a+2 b+3 c+d)\) ! \(\therefore\) Total number of arrangements \(=\frac{(\mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}+\mathrm{d}) !}{\mathrm{a} !(\mathrm{b} !)^2(\mathrm{c} !)^3}\)
COMEDK-2018
Permutation and Combination
119236
Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the number of heads is
1 20
2 9
3 120
4 40
Explanation:
A Here, the required number of ways is the number of ways in which the letters HHHTTT can be arranged. ( \(\because \mathrm{H}=\) heads, \(\mathrm{T}=\) tails \()\) Hence, they can be all together arranged in 6 ! ways but, we have to account for 3 repetitive heads and tails. Thus, the required permutation is \(=\frac{6 !}{3 ! 3 !}={ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 !}=20\)
VITEEE-2019
Permutation and Combination
119240
There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent is \(\qquad\)
1 120
2 \(89 .(8 !)\)
3 56
4 \(42 \times 5^4\)
Explanation:
C : First arrange 7 identical white ball in row is \(\frac{7 !}{7 !}=1\) way. Now for black balls we have 8 places from 8 places we can choose any of 3 places \(={ }^8 \mathrm{C}_3\) and we can arrange them in 3 ! ways as all the 3 balls are identical so it should be \(\frac{3 !}{3 !}\) hence, solution should be \(\frac{7 !}{7 !} \times{ }^8 \mathrm{C}_3 \times \frac{3 !}{3 !}=56\)
