119270
Eight different letters of an alphabet are given. Words of four letters from these are formed. The number of such words with at least one letter repeated is
C : The number of words of 4 letters with 8 different letter is \(8^4=4096\) Number of words of 4 letters with at least one letter without repetition \(={ }^8 \mathrm{P}_4=1680\) \(\therefore\) Number of 4 letter words with at least one letter repeated is \(8^4-{ }^8 \mathrm{P}_4=4096-1680=2416\)
BCECE-2008
Permutation and Combination
119291
The number of positive divisors of 4896 is
1 32
2 34
3 36
4 38
Explanation:
C factor of 4896 \(=2^5 \times 3^2 \times 17\) \(\text { Number of divisor is }\) \((1+5)(1+2)(1+1)\) \(\quad=6 \times 3 \times 2\) \(\quad=36\) Number of divisor is
COMEDK-2012
Permutation and Combination
119292
The last digit of \(583 !+7^{291}\) is
1 1
2 2
3 0
4 3
Explanation:
D We have, \(5 !=120\) \(\therefore\) If \(\mathrm{n} \geq 5\), the digit at unit place in \(\mathrm{n}\) ! is 0 Also \(7^{291}=\left(7^4\right)^{72} \cdot(7)^3=\left(7^4\right)^{72} \cdot(343)\) The digit at unit place in \(\left(7^4\right)^{72}=1\) i.e., in \(7^{291}\) is 3 The digit at unit place in \(583 !+7^{291}=0+3=3\)
COMEDK-2012
Permutation and Combination
119315
If \((n+2) !=2550 \times n\) !, then the value of \(n\) is equal to
119270
Eight different letters of an alphabet are given. Words of four letters from these are formed. The number of such words with at least one letter repeated is
C : The number of words of 4 letters with 8 different letter is \(8^4=4096\) Number of words of 4 letters with at least one letter without repetition \(={ }^8 \mathrm{P}_4=1680\) \(\therefore\) Number of 4 letter words with at least one letter repeated is \(8^4-{ }^8 \mathrm{P}_4=4096-1680=2416\)
BCECE-2008
Permutation and Combination
119291
The number of positive divisors of 4896 is
1 32
2 34
3 36
4 38
Explanation:
C factor of 4896 \(=2^5 \times 3^2 \times 17\) \(\text { Number of divisor is }\) \((1+5)(1+2)(1+1)\) \(\quad=6 \times 3 \times 2\) \(\quad=36\) Number of divisor is
COMEDK-2012
Permutation and Combination
119292
The last digit of \(583 !+7^{291}\) is
1 1
2 2
3 0
4 3
Explanation:
D We have, \(5 !=120\) \(\therefore\) If \(\mathrm{n} \geq 5\), the digit at unit place in \(\mathrm{n}\) ! is 0 Also \(7^{291}=\left(7^4\right)^{72} \cdot(7)^3=\left(7^4\right)^{72} \cdot(343)\) The digit at unit place in \(\left(7^4\right)^{72}=1\) i.e., in \(7^{291}\) is 3 The digit at unit place in \(583 !+7^{291}=0+3=3\)
COMEDK-2012
Permutation and Combination
119315
If \((n+2) !=2550 \times n\) !, then the value of \(n\) is equal to
119270
Eight different letters of an alphabet are given. Words of four letters from these are formed. The number of such words with at least one letter repeated is
C : The number of words of 4 letters with 8 different letter is \(8^4=4096\) Number of words of 4 letters with at least one letter without repetition \(={ }^8 \mathrm{P}_4=1680\) \(\therefore\) Number of 4 letter words with at least one letter repeated is \(8^4-{ }^8 \mathrm{P}_4=4096-1680=2416\)
BCECE-2008
Permutation and Combination
119291
The number of positive divisors of 4896 is
1 32
2 34
3 36
4 38
Explanation:
C factor of 4896 \(=2^5 \times 3^2 \times 17\) \(\text { Number of divisor is }\) \((1+5)(1+2)(1+1)\) \(\quad=6 \times 3 \times 2\) \(\quad=36\) Number of divisor is
COMEDK-2012
Permutation and Combination
119292
The last digit of \(583 !+7^{291}\) is
1 1
2 2
3 0
4 3
Explanation:
D We have, \(5 !=120\) \(\therefore\) If \(\mathrm{n} \geq 5\), the digit at unit place in \(\mathrm{n}\) ! is 0 Also \(7^{291}=\left(7^4\right)^{72} \cdot(7)^3=\left(7^4\right)^{72} \cdot(343)\) The digit at unit place in \(\left(7^4\right)^{72}=1\) i.e., in \(7^{291}\) is 3 The digit at unit place in \(583 !+7^{291}=0+3=3\)
COMEDK-2012
Permutation and Combination
119315
If \((n+2) !=2550 \times n\) !, then the value of \(n\) is equal to
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Permutation and Combination
119270
Eight different letters of an alphabet are given. Words of four letters from these are formed. The number of such words with at least one letter repeated is
C : The number of words of 4 letters with 8 different letter is \(8^4=4096\) Number of words of 4 letters with at least one letter without repetition \(={ }^8 \mathrm{P}_4=1680\) \(\therefore\) Number of 4 letter words with at least one letter repeated is \(8^4-{ }^8 \mathrm{P}_4=4096-1680=2416\)
BCECE-2008
Permutation and Combination
119291
The number of positive divisors of 4896 is
1 32
2 34
3 36
4 38
Explanation:
C factor of 4896 \(=2^5 \times 3^2 \times 17\) \(\text { Number of divisor is }\) \((1+5)(1+2)(1+1)\) \(\quad=6 \times 3 \times 2\) \(\quad=36\) Number of divisor is
COMEDK-2012
Permutation and Combination
119292
The last digit of \(583 !+7^{291}\) is
1 1
2 2
3 0
4 3
Explanation:
D We have, \(5 !=120\) \(\therefore\) If \(\mathrm{n} \geq 5\), the digit at unit place in \(\mathrm{n}\) ! is 0 Also \(7^{291}=\left(7^4\right)^{72} \cdot(7)^3=\left(7^4\right)^{72} \cdot(343)\) The digit at unit place in \(\left(7^4\right)^{72}=1\) i.e., in \(7^{291}\) is 3 The digit at unit place in \(583 !+7^{291}=0+3=3\)
COMEDK-2012
Permutation and Combination
119315
If \((n+2) !=2550 \times n\) !, then the value of \(n\) is equal to
119270
Eight different letters of an alphabet are given. Words of four letters from these are formed. The number of such words with at least one letter repeated is
C : The number of words of 4 letters with 8 different letter is \(8^4=4096\) Number of words of 4 letters with at least one letter without repetition \(={ }^8 \mathrm{P}_4=1680\) \(\therefore\) Number of 4 letter words with at least one letter repeated is \(8^4-{ }^8 \mathrm{P}_4=4096-1680=2416\)
BCECE-2008
Permutation and Combination
119291
The number of positive divisors of 4896 is
1 32
2 34
3 36
4 38
Explanation:
C factor of 4896 \(=2^5 \times 3^2 \times 17\) \(\text { Number of divisor is }\) \((1+5)(1+2)(1+1)\) \(\quad=6 \times 3 \times 2\) \(\quad=36\) Number of divisor is
COMEDK-2012
Permutation and Combination
119292
The last digit of \(583 !+7^{291}\) is
1 1
2 2
3 0
4 3
Explanation:
D We have, \(5 !=120\) \(\therefore\) If \(\mathrm{n} \geq 5\), the digit at unit place in \(\mathrm{n}\) ! is 0 Also \(7^{291}=\left(7^4\right)^{72} \cdot(7)^3=\left(7^4\right)^{72} \cdot(343)\) The digit at unit place in \(\left(7^4\right)^{72}=1\) i.e., in \(7^{291}\) is 3 The digit at unit place in \(583 !+7^{291}=0+3=3\)
COMEDK-2012
Permutation and Combination
119315
If \((n+2) !=2550 \times n\) !, then the value of \(n\) is equal to