A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they

Exercise 6.3 · Q7

Q. A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that

(i) all the sisters sit together

(ii) no two sisters sit together

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Use the block method for "together" and the gap method for "no two together".

^nP_r=\dfrac{n!}{(n-r)!}

(i) All 4 sisters sit together

Treat the 4 sisters as a single block. Now we are arranging $6$ brothers $+\ 1$ block $=7$ entities in a row: $7!$ ways.
Within the block, the 4 sisters can be permuted among themselves: $4!$ ways.
Total $=7!\times4! = 5040\times24$ .
Compute: $5040\times24=120960$ .

(ii) No two sisters sit together

First arrange the 6 brothers in a row: $6!=720$ ways. This creates $7$ gaps (one before, one after, and one between each pair of brothers). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers, all digits of which are odd is:$
(a) $5^6$
(b) $25$
(c) $625$
(d) $5^5$
›Reveal solutionSolution

Each of the 5 digit-positions is filled independently from the 5 odd digits.

The odd digits available are $\{1,3,5,7,9\}$ , a set of 5 digits. Since none of them is $0$ , there is no restriction on the leading digit either. Each of the 5 positions of the number can be filled in 5 ways, independ …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers all digits of which are odd is:$
(a) $5^6$
(b) 25
(c) 625
(d) $5^5$
›Reveal solutionSolution

With 5 choices of odd digit for each of the 5 positions (no restriction, since all odd digits are already non-zero), the count is $5^5$ .

The odd digits are $1,3,5,7,9$ -- exactly 5 of them.

A 5-digit number has 5 positions, and unlike a general 5-digit number problem, there's no separate restriction on the leading digit here, because none of the allowed digits (odd digits) is zero anyway.

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The number of five digit numbers in which all digits are even, is:$
(a) $4\times 5^4$
(b) $4\times 5^5$
(c) $5^5$
(d) $5\times 5$
›Reveal solutionSolution

With digits restricted to $\{0,2,4,6,8\}$ , the first digit excludes 0 (4 choices) and the remaining 4 digits are unrestricted (5 choices each), giving $4\times 5^4$ .

The even digits available are $0, 2, 4, 6, 8$ — 5 digits in total, and repetition is allowed (the question only restricts which digits may be used, not that they be distinct).

For a 5-digit number, the first (leftmost) digit cannot be 0, so it has $5-1=4$ choices.

…

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Q.A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that

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Q.A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that

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